For some reason I do not feel this is correct. Can someone please verify?
Let $X$ be compact and $Y \subset \mathbb{R}^n$ be closed. Let $f\colon X\times Y\rightarrow\mathbb{R}$ be a continuous function (under the product metric). Then, $y\mapsto\sup_{x\in X}f(x,y)$ is continuous.
Proof: Fix $y\in Y$ and $\epsilon>0$. Let $B(y)\subset Y$ be the unit ball around $y$. Since the restriction $$ f|_{X\times(\overline{B(y)} \cap Y) } $$ is uniformly continuous, we can find $0<\delta<1$ such that for all $(x_{1},y_{1}),(x_{2},y_{2})\in X\times(\overline{B(y)}\cap Y)$ satisfying $d((x_{1},y_{1}),(x_{2},y_{2}))<\delta$, $|f(x_{1},y_{1})-f(x_{2},y_{2})|<\epsilon$. Therefore, for $y^{\prime}\in\overline{B(y)}\cap Y$ such that $d(y,y^{\prime})<\delta$, $$ |\sup_{x\in X}f(x,y)-\sup_{x\in X}f(x,y^{\prime})|\leq\sup_{x\in X}|f(x,y)-f(x,y^{\prime})|=|f(x^{\star},y)-f(x^{\star},y^{\prime})|<\epsilon. $$