Above is my question. I have completed the question, but I'm not 100% about my proof for the final part - it seems like I haven't done enough.
I've shown that if $U$ and $V$ are compact, then so is $U+V$ (that's not the problem). So then, since in the second art of the question I have shown uniform approximation, it is then sufficient to show that the operator $S$ defined by $$Sf(x) = \int_0^1 f(y)A(x)B(y) dy = A(x) \int_0^1 f(y)B(y) dy$$ is compact, for continuous functions $A$ and $B$. (Now comes the bit that I'm not sure on.) This follows immediately from the second form of $Sf$ above, in particular that $Sf(x)$ is always a multiple of $A(x)$, where $A$ is fixed. So $rank(S) = 1 < \infty$, and so $S$ is compact. The result follows.
Note: This final result can be proved pretty easily using Arzela-Ascoli - please no references to that: this is a past exam question, and says "hence"! :)