Continuity of improper integral with a continuous integrand. I am a newbie in analysis and am trying to wrap my head around some continuity/compactness/finiteness concepts.
Let $f(x,y):\mathbb{R}^2\mapsto\mathbb{R}$ be a continuous function in both $x$ and $y$ and define 
$$
F(x) = \int_0^{\infty}f(x,y)dy.
$$
Assume that $F(x)<\infty$ for all $x>0$. Can I show that $F(x)$ is continuous for $x>0$? 
(The original question is different (see details below), but this is the most important part of the proof (thanks to the hint given by Brian))
Given a compact set $\mathcal{X}$ and suppose $0$ is not a member of $\mathcal{X}$. Can I show $\sup_{x\in\mathcal{X}} F(x) <\infty $?
Here are what I tried so far:


*

*Since $\mathcal{X}$ is compact so $\sup_{x\in\mathcal{X}} F(x) = \max_{x\in\mathcal{X}} F(x)$. Moreover, since $F(x)<\infty$ for any $x>0$ and $0\notin \mathcal{X}$, then $\max_{x\in\mathcal{X}} F(x)<\infty$. 
(This proof seems okay to me. But the continuity of $f(x,y)$ is not used anywhere, which makes me a little puzzled).

*Since $f(x,y)$ is continuous function and $F(x)<\infty$ for all $x>0$, then $F(x)$ is a continuous function for any $x>0$ (is this true?). Since $\mathcal{X}$ is compact so $\sup_{x\in\mathcal{X}} F(x) = \max_{x\in\mathcal{X}} F(x)$. The maximization is finite because it is the maximum of a continuous function in a compact set. (This proof does use all given conditions, but the first assessment is not clear to me. Any reference to existing theorem is appreciated.)
Any help is highly appreciated! 

For part 1, only continuous functions are guaranteed to attain their maximum on compact sets.
