Let $X$ and $Y$ be topological spaces. Wikipedia states that if $Y$ is $T_{3½}$, then so is $Y^X$. Yet I couldn't find the proof anywhere. So here's my own attempt of a proof:
(I also had had a hard time proving that if $Y$ is $T_3$, then so is $Y^X$. My professor verified that proof. Thanks, Mr. Iggy.)
Let $f \in Y^X$, and let $S(C,U)$ be a subbasis element having $f$. I shall separate $f$ from $Y^X \setminus S(C,U)$ by a continuous function $g : Y^X → [0,1]$. There are few steps for finding such $g$.
Step 1. I shall find a continuous function $h : f[C] × Y → [0,1]$ that separates the diagonal $\Delta = \{(y,y) : y \in f[C]\}$ from $f[C] × (Y \setminus U)$. Since $f[C] × f[C]$ is a compact Hausdorff space which $\Delta$ is closed within, $\Delta$ is a compact subspace of the $T_{3½}$ space $f[C] × Y$. So such $h$ exists. (Munkres' Topology, Section 33, Exercise 8.)
Step 2. Now give $g(\phi) = \sup\{h(f(x),\phi(x)) : x \in C\}$.
Is this proof correct so far? It should be fairly easy to replace $S(C,U)$ by an arbitrary basis element and then by an arbitrary open set.