The problem is here:
If any $x\in X$ had for each of its neighborhood $U$ infinitely many $n$ for which $x_n \in U$, then we could define a convergent subsequence of $(x_n)$, contradicting our assumption. (Presumably this is done by choosing for each neighborhood a sufficiently-large-indexed term in that neighborhood.)
In general topological spaces this only implies that we are able to construct a convergent net, not a convergent sequence. (A point $x$ is an accumulation point of a subset $S$ $\Leftrightarrow$ there exists a net of points of $S\setminus\{x\}$ converging to $x$.)
If $X$ is first countable at $x$ (the point $x$ has a countable base), then a sequence can be constructed. (This is more-or-less standard. We first construct a decreasing base $U_n$ at $x$ and then choose a point from each $U_n$). In particular, this works for metric spaces. Note that Rudin works only with compact subset of metric spaces in that chapter.
It is true that the one-point compactification of the rationals with their usual topology is not second countable: it is not first countable at the new point. Unfortunately, neither of the proofs that come to mind is completely elementary. I’ll sketch the simpler one. Let $\Bbb Q^*$ be the one-point compactification of $\Bbb Q$ with the usual topology, and let $p$ be the point at infinity.
I’ll show first that $\Bbb Q^*$ is a $US$-space, meaning that no sequence in $\Bbb Q^*$ converges to more than one point. Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb Q$ that converges in $\Bbb Q^*$. If $\sigma$ converges to some $x\in\Bbb Q$, the set $K=\{x\}\cup\{x_n:n\in\Bbb N\}$ is a compact subset of $\Bbb Q$, $\Bbb Q^*\setminus K$ is therefore an open nbhd of $p$ disjoint from $K$, and $\sigma$ does not converge to $p$. The subspace $\Bbb Q$ is Hausdorff, and all Hausdorff spaces are $US$-spaces, so $\sigma$ does not converge to any point of $\Bbb Q$ other than $x$, and $x$ is therefore the unique limit of $\sigma$. If $\sigma$ does not converge to any point of $\Bbb Q$, then of course its unique limit must be $p$. Thus, $\Bbb Q^*$ is a $US$-space.
$\Bbb Q^*$ is not Hausdorff: if $x\in\Bbb Q$, $x$ and $p$ cannot be separated by disjoint open sets. For suppose that $U$ is an open nbhd of $p$. Then $K=\Bbb Q^*\setminus U$ is a compact subset of $\Bbb Q$ and therefore nowhere dense in $\Bbb Q$, so $K$ cannot contain an open nbhd of $x$.
Finally, every first countable $US$-space is Hausdorff, so $\Bbb Q^*$ cannot be first countable. To see this, suppose that $X$ is a first countable $US$-space that is not Hausdorff, and let $x,y\in X$ be such that $U\cap V\ne\varnothing$ whenever $U$ and $V$ are open nbhds of $x$ and $y$, respectively. Suppose that $\{B_n:n\in\Bbb N\}$ is a countable open base at $x$, and $\{W_n:n\in\Bbb N\}$ is a countable open base at $y$. For each $n\in\Bbb N$ let $x_n\in B_n\cap W_n$; then $\langle x_n:n\in\Bbb N\rangle$ converges to both $x$ and $y$, contradicting the hypothesis that $X$ is a $US$-space. Thus, $X$ must be Hausdorff.
There are much easier examples of countable spaces that are not second countable. One is the Arens-Fort space; given a countable family $\mathscr{B}$ of open nbhds of the one non-isolated point of the space, it’s not hard to construct an open nbhd of that point that doesn’t contain any member of $\mathscr{B}$. If you know a little about ultrafilters, another fairly simple example is to let $p$ be a free ultrafilter on $\Bbb N$ and let $X=\{p\}\cup\Bbb N$. $\Bbb N$ has the discrete topology, and a local base at $p$ is $\big\{\{p\}\cup U:U\in p\big\}$. This is a subspace of the Katětov extension of $\Bbb N$, and I proved in this answer that it’s not first countable at $p$.
In this answer I described a couple more examples, one based on the ultrafilter space and one on the Arens-Fort space. They’re a bit more complicated, but they have some fairly nice properties: they’re zero-dimensional and Hausdorff and have no isolated points.
Best Answer
Hint: What is the topology on $\mathbb{Q}^*$? What does a neighborhood of $\infty$ look like? It's not at all what it would be in the case of $\mathbb{R}$.
Further Hint: $\mathbb{Q}$ is not locally compact, so the one point compactification is not Hausdorff.