It would be very appreciated if someone could review my solution. Thanks!
Problem:
Show that every locally compact Hausdorff space is completely regular.
Proof:
Let X be a locally compact Hausdorff space. One point sets are closed in X since X is hausdorff.
Then there is a space Y such that Y is the one point compactification of X where Y is compact hausdorff.
Let $x_0$ $\in$ X and B $\subset$ X such that $x_0$ $\notin$ B and B is closed in X. Then the open set U = X – B contains x$_0$. Then since U is open in X it is also open in Y by construction of the one point compactification Munkres gives. Then C = Y – U is a closed set in Y that contains B.
Then since Y is compact hausdorff it is also normal. Hence since the set {x$_0$} is closed since Y is hausdorff, by the Urysohn lemma we can define a continuous function f: Y -> [a,b] such that f(x$_0$) = a and f(x) = b for every x $\in$ C.
But since f is continuous we can take the continuous function g on a subspace of the domain of f. Namely, g: X -> [a,b] since X is a subspace of Y. Hence we have a continuous function g that maps g(x$_0$) = a and g(B) = b, since B $\subset$ C. We can replace [a,b] by [0,1] to satisfy the definition of completely regular.
Hence X is completely regular.
Best Answer
It follows from two basic facts:
$X$ embeds into its one-point compactification $\alpha X$, which is Hausdorff and compact and thus normal and hence Tychonoff.
A subspace of a Tychonoff space is again Tychonoff.
Your proof is almost a re-proof of the last fact. If it has been covered in your course, it's better to just quote that instead of redoing its proof in a special case.