Prob. 1, Sec. 29, in Munkres' TOPOLOGY, 2nd ed: The subspace $\mathbb{Q}$ of $\mathbb{R}$ is not locally compact 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?
 A: 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.
A: A briefer method:  A nbhd of $x$ is a set $N$ such that there exists an open set $U$ with $x\in U\subset N$. Let $x\in N\subset \Bbb Q$ where $N$ is a nbhd of $x$ in the space $\Bbb Q.$ Then $N$ is not compact.
Proof: There exists $r\in \Bbb R^+$ such that $\Bbb Q\cap (-r+x,r+x)\subset N.$ There exists $y\in (\Bbb R \setminus \Bbb Q)\cap (-r+x,r+x).$
Let $C_0=\{\Bbb Q\cap (-\infty,z): y>z\in \Bbb R\}.$ Let $C_1=C_0 \cup \{\Bbb Q\cap (y,\infty)\}.$ 
Then $\cup C_1=\Bbb Q$ (because if $q\in \Bbb Q \cap (-\infty,y)$ then  $q\in \Bbb Q \cap (-\infty,(y+q)/2 )\in C_0\subset C_1 ,$ so $q\in \cup C_1)$.
So $C_1$ is an open cover of $N$ in the space $\Bbb Q.$
Let $F$ be a finite subset of $C_1.$ Then:
(i). If $F \cap C_0=\emptyset$ then $N\cap (-r+x,y)=\Bbb Q\cap (-r+x,y)$ is a non-empty subset of $ N$ that is not covered by $F.$
(ii). If $F\cap C_0\ne \emptyset$ then (as $F$ is finite) there exists $z_0=\max \{z<y:\Bbb Q \cap (-\infty,z)\in F\}.$
Let $z_1=\max (z_0,-r+x).$ Then $-r+x\le z_1<y$ so $N\cap (z_1,y)=\Bbb Q\cap (z_1,y)$ is a non-empty subset of $N$ not covered by $F.$
