I got a bit confused about the following:
Suppose $g: X \times Y \to \mathbb{R}$ is continuous in $x$ and $y$, where $X$ and $Y$ are some metric spaces. I was assuming that then
$h : X \to \overline{\mathbb{R}}$
$h(x) = \underset{y} \sup g(x,y)$
is continuous. The idea I have in mind is the following:
Assume some open set $B$ in $\overline{\mathbb{R}}$. Take any $x$ in $h^{-1}(B)$.
If $h(x)$ is finite, we find some $y'$ where $g(x,y')$ is sufficiently close to $h(x)$ and then by continuity of $g$ some $x'$ in a neighborhood of $x$ such that again $g(x',y')$ is sufficiently close to $h(x)$.
On the other hand, if we would find for $x'$ some $y''$ such that $g(x',y'')$ is too big, we have that $g(x,y'')$ would also be too big.
For the infinite case, we argue similar to the first part of the preceding arguments (i.e. if $g(x,y)$ is unbounded, then for some neighboring $x$, $g$ should also be unbounded).
Am I misleaded here?
I don't think the statment is true as written. For example take $X = Y = \mathbb{R}_{\geq 0}$ under the standard topology and $$g(x,y) = \frac{xy}{xy + 1}$$ Then $h(0) = 0$ but for every $x > 0$, $h(x) = 1$.
However, the statment is true if we assume $Y$ is compact. Fix $x \in X$ and $\epsilon > 0$. Then for every $y \in Y$, let $U_y$ be an open rectangle containing $(x,y)$ satisfying $$\sup_{(a,b),(c,d) \in U_y} d(g(a,b), g(c,d)) < \epsilon $$. These sets form an open cover for $\{(x,y): y \in Y\}$ which is compact so there is a finite subcover $U_1, U_2, \cdots, U_n$. Since each was an open retangle, there exists a $\sigma > 0$ such that if $d(x',x) < \sigma$, then for every $y \in Y$, $$(x',y) \in \bigcup_{i = 1}^n U_i$$So if $d(x',x) < \sigma$, $$d(h(x'),h(x)) < \epsilon$$ so $h$ is continuous.