How to prove that $\mathbb{Q} \subset \mathbb{R}$ is not locally compact directly?

How to prove that $\mathbb{Q} \subset \mathbb{R}$ is not locally compact directly? That is, how to construct a cover of an arbitrary neighborhood (e.g. $[0, 1] \cap \mathbb{Q}$) that does not have a finite subcover?

Let $X:= \mathbb Q \cap [0,1]$. I'll show that if $U \subset X$ then $U$ doesn't have compact closure. Since $X$ is hausdorff this shows that it's not locally compact. It suffices to show that for any epsilon an open ball in X doesn't have compact closure. Given $x \in X$ and some $\epsilon>0$ pick an irrational $y \in B(x,\epsilon)$. Set $a=x-\epsilon$ and $b=x+\epsilon$, then the following is an open cover of $B(x,\epsilon)$ with no finite subcover:

$$\mathcal U := \left\{ \left( \left(a,y-\frac{1}{n}\right) \cup \left(y+\frac{1}{n},b\right) \right) \cap \mathbb Q \mid n \in \mathbb N\right\}.$$

To see this note that $\mathcal U$ is a nested covering and that for any $n$ it doesn't cover $B(x,\epsilon)$.

Choose irrational point $i$ inside $[0,1]$. Take the class $O = \{[0, q) \cup (r, 1] \quad: \quad(q,r) \in \mathbb Q^2 \cap(0,1)^2,\quad q < i < r\}$ as your covering. Now we have a cover for the rationals in $[0,1]$ without a finite subcover, showing that the space is not compact.

For the sake of rigor:

$O$ is an open cover in the subspace topology relative to the compact set $[0,1]$.

$\cup O = [0,1]- \{i\}$, showing that we do, in fact, cover the rationals in $[0,1]$.

Consider the union of any finite subcover: $\cup_{k=1}^N[0,q_k)\cup(r_k,1]$ where $(q_k,r_k) \in \mathbb Q^2 \cap(0,1)^2,\quad\mathrm{and}\quad q < i < r$. This set is equivalent to $[0,\mathrm{max}_k\{q_k\})\cup(\mathrm{min}_k\{r_k\},1]$, where $k$ is understood to range from $1$ to $N$.

We conclude that any finite subcover misses the points in the interval $[\mathrm{max}_k\{q_k\},\mathrm{min}_k\{r_k\}]$

In general, choose any neighborhood in $\mathbb R$. We must be able to find a segment $(a,b)$ within the neighborhood. Repeat the argument using segment $(a,b)$ instead of $[0,1]$.

We conclude that any neighborhood of $\mathbb R$ has an open cover for which no subcover can cover the rationals in that neighborhood.