Locally compact Hausdorff then Tychonoff

Question

This is my proof. If $X$ is locally compact Hausdorff then for each $x \in X$ its compact neighborhoods form a neighborhood basis at $x$. For an arbitrary closed set $C \not\ni x$ let $V \subset X \setminus C$ be a compact neighborhood of $x$. For each dyadic rational $r$, let's define open neighborhoods $U_r$ inductively s.t. $s < r \implies \overline{U_s} \subset U_r$. Let $U_0 = \operatorname{int}(V), U_1 = X - C$. $\{A \mid A \subset U_1 \text{is a compact neighborhood of a }x \in V\}$ is a cover of $V$ so there is a finite subcover $S$. Let $W = \bigcup{S}$ then it is compact because it is a finite union of compact sets. Let $U_{1/2} = \operatorname{int}(W)$. Then it is open and $\overline{U_0} \subset U_{1/2}, \overline{U_{1/2}} \subset U_1$. Apply this method analogously to define for all other dyadic rationals. Now let's define a function $f\colon X \to [0, 1]$ $$f(x) = \begin{cases} \operatorname{inf}\{r \mid x \in U_r\}& x \in U_1 \\ 1& x \not\in U_1 \end{cases}$$ Then $f(x) = 0$ and $f \equiv 1$ on $C$. Let's show $f^{-1}(\infty, a)$ and $f^{-1}(b, \infty)$ are open. If $a > 1$ it is $X$ and if $a \leq 1$, $f^{-1}(\infty, a) = \bigcup{\{U_r \mid r < a\}}$ so open. $f^{-1}(b, \infty) = \bigcup{\{X \setminus U_r \mid r > b\}} = \bigcup{\{X \setminus \overline{U_s} \mid s > b\}}$ so open. Finally $f$ is continuous so $X$ is Tychonoff. Is this proof correct?

Answer 1

It is possibly correct (you do essentially use normality like structures we get from the local compactness), but why not use the fact that the Alexandroff (one-point) compactification $\alpha(X)$ of $X$ is compact Hausdorff (and so is normal and hence Tychonoff) and contains $X$ as a subspace (and subspaces of Tychonoff spaces are Tychonoff)? Much cleaner. Why redo the proof of Urysohn's lemma for $X$ when we already have it for $\alpha(X)$? Efficiency.