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Here is my attempt to prove that $\mathbb Q$ is not locally compact. (My questions are below the proof.)

Suppose $\mathbb Q$ is locally compact. Then it is locally compact at every point. Let $x\in \mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $\mathbb Q$ and a neighborhood $U$ of $x$ in $\mathbb Q$ such that $x\in U\subseteq K\subseteq Q$. (The neighborhood $U$ is of the form $(a,b)\cap \mathbb Q$ for some $a,b\in\mathbb Q$; here $x\in (a,b)\subset \mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $\overline K$. But since $K$ is a compact subspace of the Hausdorff space $\mathbb Q$, it is closed in $\mathbb Q$, so $\overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $K\subseteq Q$.

The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $\mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $\mathbb Q$.

And otherwise does the proof look okay?

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Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $\mathbb{Q}$.

As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $r\in (a,b)$. Then for any $n\in \mathbb{N}$, there exists a rational number $x_n$ such that $$ r < x_n < r+\frac{1}{n}. $$ This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n \in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_n\in (a,b)$ for all $n\in\mathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)\cap\mathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.

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