Here is Prob. 1, Sec. 29, in the book Topology by James R. Munkres, 2nd edition:
Show that the rationals $\mathbb{Q}$ are not locally compact.
My Attempt:
Here the topology on the set $\mathbb{Q}$ of rational numbers is the same as the subspace topology that $\mathbb{Q}$ inherits from the standard topology on the set $\mathbb{R}$ of real numbers.
As the standard topology on $\mathbb{R}$ has as a basis the collection of all the open intervals of the form $(a, b)$, where $a, b \in \mathbb{R}$ and $a < b$, so by virtue of Lemma 16.1 in Munkres the subspace topology on $\mathbb{Q}$ has as a basis the collection of all the intersections $(a, b) \cap \mathbb{Q}$, where $a, b \in \mathbb{R}$ and $a < b$.
Let $q$ be any point of $\mathbb{Q}$. Let us suppose that $\mathbb{Q}$ is locally compact at $q$. Then there exists a neighborhood $U$ of $q$ in $\mathbb{Q}$ and a compact subspace $C$ of $\mathbb{Q}$ such that $$ U \subset C. \tag{0} $$
As $U$ is open in $\mathbb{Q}$, so $$U = V \cap \mathbb{Q} \tag{1} $$ for some open set $V$ in $\mathbb{R}$.
Now as $q \in V$ and as $V$ is open in $\mathbb{R}$, so there exists an open interval $(a, b)$ on the real line such that $$q \in (a, b) \subset V. \tag{2}$$ Thus we have $$ a < q < b. $$ Let us choose some irrational numbers $c$ and $d$ such that $$ a < c < q < d < b. \tag{3}$$ Then by (2) above we have $$ q \in (c, d) \subset (a, b) \subset V,$$ and then by (1) above we also have $$q \, \in \, (c, d) \cap \mathbb{Q} \, \subset \, (a, b) \cap \mathbb{Q} \, \subset \, V \cap \mathbb{Q} \, = \, U, $$ that is $$ q \, \in \, (c, d) \cap \mathbb{Q} \, \subset \, U, \tag{4}$$ and also $$ (c, d) \cap \mathbb{Q} \ = \ [c, d] \cap \mathbb{Q}, \tag{5} $$ because the endpoints $c$ and $d$ either interval are not in either of the two sets involved.
Thus from (0), (4), and (5) above we also have $$q \, \in \, [c, d] \cap \mathbb{Q} \, \subset \, C. \tag{6}$$
Now $C$ is a compact subspace of $\mathbb{Q}$; moreover $\mathbb{Q}$, being a metrizable space, is also a Hausdorff space, by the discussion in the third paragraph of Sec. 21 in Munkres. So by Theorem 26.3 in Munkres $C$, being a compact subspace of the Hausdorff space $\mathbb{Q}$, is also closed in $\mathbb{Q}$. Therefore using (6) above we can also conclude that $$ [c, d] \cap \mathbb{Q} = \big( [c, d] \cap \mathbb{Q} \big) \cap C$$ is also closed in $C$.
Thus we have seen that $[c, d] \cap \mathbb{Q}$ is a closed subset of the compact space $C$. So by Theorem 26.2 in Munkres the subspace $[c, d] \cap \mathbb{Q}$ is also compact (as a subspace of $C$); but since $C$ is a subspace of $\mathbb{Q}$, therefore $[c, d] \cap \mathbb{Q} = (c, d) \cap \mathbb{Q}$ is also a compact subspace of $\mathbb{Q}$.
Therefore every open covering of $[c, d] \cap \mathbb{Q}$ has a finite sub-collection that also covers $[c, d] \cap \mathbb{Q}$.
However, as $c$ and $d$ are irrational numbers [Please refer to (3) above.], so we now show that what has stated in the preceding paragraph is not true.
In what follows, the set $\mathbb{N}$ denotes the set of all the positive integers, namely, $1, 2, 3, \ldots$.
Let us consider the collection $$ \left\{ \ \left( c + \frac{d-c}{n+2},\, d - \frac{d-c}{n+2} \right) \cap \mathbb{Q} \ \colon \ n \in \mathbb{N} \ \right\}. \tag{A} $$ This collection is an open covering of $[c, d]\cap \mathbb{Q}$ such that no finite subcollection of this collection can cover $[c, d] \cap \mathbb{Q}$; for if $$ \left\{ \ \left( c + \frac{d-c}{n_1+2}, \, d - \frac{d-c}{n_1+2} \right) \cap \mathbb{Q},\ \ldots, \ \left( c + \frac{d-c}{ n_r + 2 }, \, d - \frac{d-c}{n_r +2} \right) \cap \mathbb{Q} \ \right\} \tag{B} $$ is any finite sub-collection of the collection in (A) above, where $n_1, \ldots, n_r \in \mathbb{N}$ such that $n_1 < \cdots < n_r$, then \begin{align} & \ \ \ \bigcup_{j=1}^r \left[ \left( c + \frac{d-c}{n_j + 2}, \, d - \frac{d-c}{n_j +2} \right) \ \cap \ \mathbb{Q} \right] \\ &= \left[ \bigcup_{j=1}^r \left( c + \frac{d-c}{n_j + 2}, d - \frac{d-c}{n_j +2} \right) \right] \ \cap \ \mathbb{Q} \\ &= \left( c + \frac{d-c}{n_r + 2}, \, d - \frac{d-c}{n_r +2} \right) \ \cap \ \mathbb{Q} \\ &\subsetneqq [c, d] \cap \mathbb{Q}. \end{align} thus showing that $[c, d] \cap \mathbb{Q}$ or $(c, d) \cap \mathbb{Q}$ [Refer to (5) above.] is not compact.
Thus we have reached a contradiction. Therefore our supposition that $\mathbb{Q}$ is locally compact at $q$ is wrong. Hence $\mathbb{Q}$ is not locally compact at any point $q \in \mathbb{Q}$.
Is my proof correct in each and every detail? If so, then is my presentation clear enough too? If not, then where are the issues?
Your proof is both correct and clear.
I think that a simpler approach would consist in proving that there are sequences of elements of $C$ without convergent subsequences. That is easy, of course: you just take a sequence of elements of $C$ which converges (in $\mathbb R$) to an irrational number.