The collection of continuous functions between any 2 compact Hausdorff spaces forms a set I would like to show precisely what I have stated in the title (assuming that it is correct; I have reason to suspect it is, thanks to a tricky past exam paper I'm trying to surmount); namely, that given any 2 compact Hdf topological spaces $T_1$ and $T_2$, the class of continuous functions between them forms a set (rather than, say, a class which is too large to be a set). I suspect this result might be obvious in a much more general sense, and just wanted to check my thinking: is it actually valid to say that for _any_ function between 2 fixed sets $S_1$ and $S_2$, the function is simply a set of ordered pairs $(a,f(a))$ and therefore contained in the power set of the union $S_1 \cup S_2$, so is again a set (by standard set-theoretic axioms for power sets and unions)?

tl;dr You are correct.

One usually takes the ordered pair $(a, b)$ to be an abbreviation for the set $\\{\\{a\\}, \\{a, b\\}\\}$, so if $a\in A$ and $b\in B$, then $\def\pow#1{{\mathcal P}{(#1)}} (a,b)\in\pow{\pow{A\cup B}}$, and the cartesian product $A\times B$ is a subset of $\pow{\pow{A\cup B}}$, which is a set by the union and power set axioms.

A function $f:A\to B$ is a subset of $A\times B\subset \pow{\pow{A\cup B}}$, and so is a set and also is an element of $\pow{\pow{\pow{A\cup B}}}$. The set $\mathcal F$ of all functions $f:A\to B$ is therefore a subset of $\pow{\pow{\pow{A\cup B}}}$ and is a set. The set of all continuous functions is a subset of $\mathcal F$, and is therefore a set, an element of $\pow{\pow{\pow{\pow{A\cup B}}}}$.