To be honest, the definition in Bak and Newman is the wort definition of simple connectivity I had ever seen. I will prove its equivalence to the traditional definition below. The proof is not short and is definitely outside of the traditional complex analysis syllabus.
In what follows, I identify the 1-point compactification of $R^n$, $R^n\cup \{\infty\}$ with the sphere $S^n$.
I will use the notation $B(c,r)=\{x\in R^n: |x-c|<r\}$ for the open $r$-ball centered at $c$.
Definition. Given a closed subset $A\subset R^n$ and $\epsilon>0$, define the $\epsilon$-neighborhood of $A$ in $R^n$ as
$$
N_\epsilon(A):= \{x\in R^n: \exists y\in A, |x-y|<\epsilon\} = \bigcup_{a\in A} B(a,r).
$$
Then
$$
\bar{N}_\epsilon(A):= \{x\in R^n: \exists y\in A, |x-y|\le\epsilon\}
$$
is the closure of $N_\epsilon(A)$ in $R^n$.
Definition. I will say that a subset $A\subset R^n$ satisfies the BN (for Bak and Newman) property if for every $\epsilon>0$ and every $a\in A$, there exists a continuous map $q: [0,\infty)\to N_\epsilon(A)$ with $q(0)=a$ and
$$
\lim_{t\to\infty} q(t)=\infty,
$$
i.e.
$$
\lim_{t\to\infty} |q(t)|=\infty.
$$
Our goal is to prove
Theorem 1. The following are equivalent for open connected subsets $U\subset R^2$:
i. $U$ is simply connected in the traditional sense, i.e. every continuous map $S^1\to U$ extends to a continuous map $D^2\to U$.
ii. The complement $A:= R^2-U$ satisfies the BN property.
The proof is somewhat tricky: The direction ii$\Rightarrow$i is rather straightforward, but the converse will require some work.
I could not find any proof shorter than the one written below.
I will be using the following characterization of simply-connected open planar sets; see here or my argument here for a proof.
Theorem 2. The following are equivalent for a nonempty open connected subset $U\subset R^2$:
$U$ is simply connected in the traditional sense, i.e. every continuous map $S^1\to U$ extends to a continuous map $D^2\to U$.
$S^2-U$ is connected.
In view of this theorem, we will need to analyze closed subsets $A\subset R^2$ such that $\hat{A}:= A \cup\{\infty\}$ is connected.
The proofs will go through in all dimensions; therefore, I will be considering subsets of $R^n$, $n\ge 1$. Note that a subset $A\subset R^n$ such that $\hat{A}:= A \cup\{\infty\}$ is connected, is necessarily unbounded. Then $\hat{A}$ is the closure of $A$ in $S^n$.
Lemma 1. For a closed subset $A\subset R^n$ the following are equivalent:
$\hat{A}$ is connected.
For every open neighborhood $V$ of $A$ in $R^n$, every component $V_j$ of $V$ is either unbounded or
is disjoint from $A$.
Every connected component of $A$ is unbounded.
Proof. Suppose that $\hat{A}$ is connected but there exists an open neighborhood $V$ of $A$ in $R^n$ and a bounded
component $V_j$ of $V$ such that $V_j\cap A=A_j\ne \emptyset$. Then $A_j$ is clopen (both closed and open) in $A$ (in the subspace topology). Since $V_j$ is open in $S^n$, the subset $A_j$ is also open in $\hat{A}$ (in the subspace topology). Since $V_j$ is bounded, so is $A_j$. Hence, $\infty$ does not belong to the closure of $A_j$ in $S^n$, hence, $A_j$ is closed in $\hat{A}$. This contradicts connectedness of $\hat{A}$.
Conversely, suppose that 2 holds but $\hat{A}$ is not connected. Then, since $\hat{A}$ is compact, there exists a pair of disjoint open subsets $V_1, V_2\subset S^n$ such that $\infty\in V_1\cap \hat{A}$ and $V_2\cap A\ne \emptyset$. But then $V_2$ is an open and bounded subset of $R^n$ whose intersection with $A$ is nonempty, contradicting 2.
Let us now prove equivalence of 2 and 3. Suppose that 3 holds. Consider an open neighborhood $V$ of $A$ in $R^n$
and a connected component $V_j$ of $V$. If $V_j\cap A\ne\emptyset$ then $V_j$ contains a component $A_j$ of $A$. Since $A_j$ is unbounded, so is $V_j$.
Suppose that 2 holds, but $A$ contains a bounded connected component $A_1$. Consider the system of open $r$-neighborhoods
$N_r(A)$ for $r=1/i, i\in {\mathbb N}$. Then
$$
\bigcap_{r} N_r(A)= \bigcap_{r} \bar{N}_r(A)=A.
$$
For each $i$ there exists a connected component $V_i$ of $N_{1/i}(A)$ containing the unbounded component $A_1$ of $A$. I assume that $A_1$ is contained in an open ball $B(R)$ of radius $R$. For each $i$, the intersection of $\bar{V}_i$ (the closure of $V_i$ in $R^n$) with the boundary sphere $S(R)$ of the ball $B(R)$ is compact and nonempty. These intersections form a nested sequence of compact nonempty subsets:
$$
\bar{V}_i\cap S(R) \subset \bar{V}_{i-1}\cap S(R).
$$
Therefore, their intersection $K$ is nonempty and is disjoint from $A_1$. Hence,
$$
C:= \bigcap_{i} \bar{V}_i
$$
is strictly larger than $A_1$. At the same time, since $\bar{V}_i\subset \bar{N}_{1/i}(A)$, $C$ is contained in $A$.
Lastly, since each $\bar{V}_i$ is connected (as the closure of a connected subset $V_i$) their intersection $C$ is also connected.
Thus, $A_1$ is not a maximal connected subset of $A$, which is a contradiction. qed
Lemma 2. Suppose that $A\subset R^n$ is a closed subset which satisfies the BN property.
Then the set $\hat{A}$ is connected.
Proof. Assume that $\hat{A}= A \cup\{\infty\}$ is not connected. Then, in view of Lemma 1, there exists an open neighborhood $V$ of $A$ in $R^n$ and a bounded component $V_1$ of $V$ which has nonempty intersection $A_1=V_1\cap A$ with $A$. Since
$A_1=\bar{V}_1\cap A$ is closed and bounded, it is compact; hence, there exists an $\epsilon>0$ such that $N_\epsilon(A_1)\subset V_1$. For every path $p: [0,\infty)\to N_\epsilon(A)$ such that $p(0)\in A_1$, the image of $p$ is contained in $V_1$, i.e. is bounded. This contradicts the BN property. qed
The remainder is a proof of the converse to Lemma 2 (Lemma 7); the proof is substantially more difficult that the proof of the lemma.
First, I will need some technical results. Much of this is quite standard (in some areas of topology) but would be quite out of place in a course in Complex Analysis.
Definition. A continuous map $f: X\to Y$ between two metric spaces is called metrically proper if preimages of bounded subsets under $f$ are bounded.
Remark. If $X, Y$ satisfy the Heine-Borel property (closed and bounded subsets are compact), then a continuous map $f: X\to Y$
is metrically proper if and only if it is proper, i.e. preimages of compact sets are compact.
The following lemma is straightforward, I will omit the proof:
Lemma 3. A continuous map $q: X\to Y$ is metrically proper if and only if it sends sequences diverging to infinity to sequences diverging to infinity, i.e. for every sequence $x_i\in X$ such that
$d(x_1, x_i)\to\infty$, it follows that
$$
\lim_{i\to\infty} d(f(x_1), f(x_i))=\infty.
$$
Corollary. A continuous map $q: [0,\infty)\to R^n$ is proper if and only if
$$
\lim_{t\to\infty} |q(t)|=\infty.
$$
I will be using metrically proper maps between graphs and $R^n$.
A graph $G$ is said to have finite valence if the number of edges incident to each vertex is finite (not necessarily uniformly bounded). Let $G$ be a connected graph. I will equip the graph $G$ with the graph-metric $d_G$ where every edge of $G$ has unit length (and is isometric to the init interval) and the distance between any two vertices is the length of the shortest edge-path between them. Thus, connected graphs become metric spaces.
Remark. A connected graph $G$ of finite valence, equipped with the graph metric $d_G$ as above, satisfies the Heine-Borel property.
Lemma 4. Suppose that $G$ is connected, has finite valence and is unbounded, i.e. has infinite diameter. Then $G$ contains a proper ray, i.e. there exists a (metrically) proper map $p: [0,\infty)\to G$ sending integers to vertices of $G$ and linear on every
interval $[n, n+1]$ (with the image equal to the edge spanned by $p(n), p(n+1)$).
Proof. Fix a vertex $u\in G$ and consider a sequence of vertices $v_i\in G$ such that $d_G(u,v_i)\to\infty$. For each $i$
let $p_i$ denote a shortest path from $u$ to $v_i$. Then, using finite valence of $G$ prove that the sequence of paths $p_i$ has a convergent subsequence whose limit is
an injective map $p: [0,\infty)\to G$ sending integers to vertices of $G$ and linear on every
interval $[n, n+1]$. Then verify that $p$ is (metrically) proper. qed
Lemma 5. Fix a positive number $\delta>0$.
Suppose that $G$ is a connected graph and $f: G\to R^n$ is a map which sends each edge of $G$ linearly
to a line segment. (I will call such maps "linear".) Assume that for any two distinct vertices $u, v$ of $G$,
$|f(u)- f(v)|\ge \delta$. Then the map $f$ is (metrically) proper.
Proof. First observe that for every bounded subset $B\subset R^n$ the preimage $f^{-1}(B)$ contains only finite number of vertices
(because images of vertices of $G$ are $\delta$-separated). From this conclude that $f$ is (metrically) proper. qed
Given a subset $C\subset R^n$ and a number $r>0$ define a graph $G=G(C,r)$, whose vertex set equals $C$ and two vertices $u, v$ are connected by an edge if and only if $|u-v|< r$. This graph comes equipped with a linear map $f: G\to R^n$ which sends
each vertex of $G$ to the corresponding point in $R^n$.
Suppose that $A\subset R^n$ is a closed connected subset; fix $r>0$. Define $C\subset A$ to be a
maximal subset of $A$ such that no two distinct points $x, y\in A$ satisfy $|x-y|<r$ ($C$ is $r$-separated).
Then, in view of the maximality
of $C$,
$$
A\subset V= \bigcup_{c\in C} B(c,r).
$$
Define the graph $G=G(C,2r)$ and let $f: G\to R^n$ denote the associated linear map.
This graph $G$ has finite valence, since its vertices are $r$-separated in $R^n$.
For every edge $e=[u,v]$ of $G$,
$f(e)\subset B(u,r)\cup B(v,r)\subset V$. Hence,
the image of $f$ is contained in the open neighborhood $V$ of $A$.
Lemma 6. The graph $G$ is connected.
Proof. Suppose that $G$ is disconnected. Then the vertex set of $G$ splits as the disjoint union $C_1\sqcup C_2$ such that no vertex of $C_1$ is connected by an edge to $C_2$. Hence, the balls $B(c_1,r), B(c_2,r)$ are disjoint whenever $c_1\in C_1, c_2\in C_2$. Then
$$
A\subset \bigcup_{c\in C_1} B(c,r) \sqcup \bigcup_{c\in C_2} B(c,r) =V,
$$
contradicting connectedness of $A$. qed
Notice that if $A$ is unbounded, $C$ is unbounded as well, hence, the graph $G$ has infinitely many vertices and, thus, is unbounded too.
At last we can prove
Lemma 7. Suppose that $A\subset R^n$ is a closed subset such that every connected component of $A$ is unbounded. Then $A$ satisfies the BN property.
Proof. Take $a\in A$ and let $A_1$ be the connected component of $a$ in $A$. Then $A_1$ is a closed connected subset of
$R^n$. Take $r=\epsilon$ and construct a maximal $r$-separated subset $C\subset A_1$ containing $a$ and
the connected graph $G=G(C,r)$ as above. By Lemma 4, $G$ contains a proper ray $p: [0,\infty)\to G$. By Lemma 5, the map $f: G\to R^n$ is proper, hence, the composition $q= f\circ p$ is a proper map $[0,\infty)\to R^n$. The image of this map is entirely contained in
$V=N_r(C)\subset N_r(A)$. Hence, $A$ satisfies the BN property. qed
This concludes the proof of Theorem 1.
