Is the interlocking interval topology path connected?

Question

Let $X=\{x\in\mathbb R:x>0,x\notin\mathbb N\}$ and define $\mathcal B=\{(0,\frac{1}{n}):n\in\mathbb N,n\ge 2\}\cup\{(0,\frac{1}{n}))\cup(n,n+1):n\in\mathbb N,n\ge 1\}$. It can be shown that $\mathcal B$ is a basis for a topology $\tau$ on $X$. Is the topological space $(X,\tau)$ path-connected?

Edit:
I came across this question in an exam, I found it quite challenging and after some hours of attempts I started searching online for a solution. Then I saw that not even "Counterexamples in topology" by Lynn Arthur Steen & J. Arthur Seebach says anything about path-connectedness of this space, situated at counterexample #54. The book however made me find out that a subbasis for $(X,\tau)$ is provided by $\{(0,\frac{1}{n}))\cup(n,n+1):n\in\mathbb N,n\ge 1\}$. Feel free to use this fact, even though I'd rather not to, given how the question was originally formulated.

Edit 2:
Inspired by the top answer, I attempted proving that $X$ is locally path-connected. Therefore, I tried to show that every set of the form $(0,\frac{1}{n})$ and $(0,\frac{1}{n})\cup(n,n+1)$ is path-connected. As the answer suggests, $(0,\frac{1}{n})$ is path-connected because it would be with the Euclidean topology. If we take $x,y$ in $(0,\frac{1}{n})\cup(n,n+1)$, we have three cases: if both $x,y$ lie in $(0,\frac{1}{n})$ we are done, since we've already shown there's a continuous path connecting x and y in $(0,\frac{1}{n})$. If both $x,y$ lie in $(n,n+1)$ we are also done, since $(n,n+1)$ has the indiscrete topology. The toughest part is when $x\in(0,\frac{1}{n})$ and $y\in(n,n+1)$. Following the top answer's argument, I defined $\gamma\colon[0,1]\longrightarrow(0,\frac{1}{n})\cup(n,n+1)$ by setting $\gamma(t)=x$ $\forall t\in[0,\frac{1}{2})$, $\gamma(\frac{1}{2})=\frac{1}{n+1}$ and $\gamma(t)=y$ $\forall t\in(\frac{1}{2},1]$. This should be a continuous path connecting $x$ and $y$.
I already know that $X$ is connected (it is even hyperconnected, as every two nonempty open sets must intersect in some $(0,\frac{1}{n})$), and finally we can use the fact that a connected, locally path-connected topological space is path-connected.

Answer 1

Weirdly enough, I think the answer to your question is yes. The proof can be seperated in two claims:

1. The interval $(0,1)$ with the subspace topology induced from the interlocking interval topology is path-connected. Indeed, for a path connecting $a$ and $b$ in that interval, one can simply take the identity function on the closed interval $[a,b]$. Since the topology induced by the i. i. t. is coarser than the Euclidean topology, the preimage of every open set is then trivially open.

2. For every $a \in (n, n+1)$ with $n \in \mathbb{N}_{\geq 1}$, the two-element set $\{a, \frac{1}{n+1}\}$ with the subspace topology is path-connected. Indeed, it is easy to see that every open set containing $a$ must contain $\frac{1}{n+1}$, but that the converse is not true. Therefore, this two-element subspace contains three open sets: $\emptyset$, $\{\frac{1}{n+1}\}$ and $\{a, \frac{1}{n+1}\}$. One can then check that the function $f:[0,1] \longrightarrow \{a, \frac{1}{n+1}\}$ defined by $f(0) = a$ and $f(x) = \frac{1}{n+1}$ for $x \neq a$ is continuous.

The whole statement then follows by concatenating paths.