Since for me $\Bbb N$ includes $0$, which is inconsistent with the defintion of $d$, I’ve written $\Bbb Z^+$ instead.
For the triangle inequality, let $x=\langle x_k:k\in\Bbb Z^+\rangle$ and $y=\langle y_k:k\in\Bbb Z^+\rangle$ be distinct points of $A$, and suppose that $d(x,y)=\frac1n$. Let $z=\langle z_k:k\in\Bbb Z^+\rangle\in A$; we want to show that
$$d(x,z)+d(z,y)\ge\frac1n\;.\tag{1}$$
If not, then $d(x,z)<\frac1n$ and $d(z,y)<\frac1n$, which means that $x_k=z_k=y_k$ for $k\le n$. But then $d(x,y)\le\frac1{n+1}$, contradicting the original assumption. Thus, at least one of $d(x,z)$ and $d(z,y)$ is $\frac1n$ of more, and $(1)$ holds.
For each $n\in\Bbb Z^+$ let $x^{(n)}=\langle x_k^{(n)}:k\in\Bbb Z^+\rangle\in A$, let $\sigma=\langle x^{(n)}:n\in\Bbb Z^+\rangle$, and suppose that $\sigma$ is a Cauchy sequence in $A$; we want to show that $\sigma$ converges to some point of $A$. Since $\sigma$ is Cauchy, for each $\epsilon>0$ there is an $r_\epsilon\in\Bbb Z^+$ such that $d(x^{(m)},x^{(n)})<\epsilon$ whenever $m,n\ge r_\epsilon$. In particular, for each $s\in\Bbb Z^+$ there is an $r_{1/s}\in\Bbb Z^+$ such that $d(x^{(m)},x^{(n)})<\frac1s$ whenever $m,n\ge r_{1/s}$. But $d(x^{(m)},x^{(n)})<\frac1s$ if and only if $x_k^{(m)}=x_k^{(n)}$ for each $k\le s$, so we’re saying that
for each $s\in\Bbb Z^+$ there is an $r_{1/s}\in\Bbb Z^+$ such that whenever $m,n\ge r_{1/s}$, then $x_k^{(m)}=x_k^{(n)}$ for each $k\le s$.
Show that this implies that for each $k\in\Bbb Z^+$, the sequence $\langle x_k^{(n)}:n\in\Bbb Z^+\rangle$ is eventually constant: there are some $x_k\in\Bbb Z^+$ and some $m_k\in\Bbb Z^+$ such that $x_k^{(n)}=x_k$ for all $n\ge m_k$.
Then let $x=\langle x_k:k\in\Bbb Z^+\rangle$, and show that $\sigma$ converges to $x$.
For part (b), suppose that $A=\bigcup_{n\in\Bbb Z^+}K_n$, where each $K_n$ is closed. Because $A$ is a complete metric space, it’s a Baire space, so there must be an $n\in\Bbb Z^+$ such that $K_n$ has non-empty interior.
Show that there is a finite sequence $\langle a_1,\ldots,a_m\rangle$ of positive integers such that $$B=\{\langle x_k:k\in\Bbb Z^+\rangle\in A:x_k=a_k\text{ for }k\le m\}$$ is a clopen subset of $K_n$.
Show that $B$ is not compact. (In thinking about this you may or may not find it helpful to notice that $B$ is homeomorphic to $A$: it’s really just $\{\langle a_1,\ldots,a_m\rangle\}\times A$, where $\langle a_1,\ldots,a_m\rangle\in(\Bbb Z^+)^m$.)
Conclude that $K_n$ is not compact.
Remark: It’s hardly obvious, but as a topological space $A$ is homeomorphic to $\Bbb R\setminus\Bbb Q$ with the topology that it inherits from $\Bbb R$, though $d$ is very different from the Euclidean metric (since the latter is not a complete metric on $\Bbb R\setminus\Bbb Q$).
