Is $f(x)=\sup_{y\in K}g(x, y)$ a continuous function? Let $K\subset \mathbb R^n$ be a compact subset and consider a continuous function $g:K\times K\longrightarrow \mathbb R$. Define $f:K\longrightarrow \mathbb R$ by, $$f(x)=\sup_{y\in K}g(x, y).$$ Is $f$ a continuous function?

Yes. Since $K$ is compact, $K\times K$ is also compact. Since $g$ is continuous, $g$ is uniformly continuous.
Let $x_0\in K$. Given $\epsilon>0$ there is a $\delta>0$ such that
$$|x-x_0|<\delta\implies |g(x_0,y)-g(x,y)|<\epsilon\quad\forall y\in K.$$
Then
$$
g(x_0,y)\le g(x,y)+\epsilon\le f(x)+\epsilon\quad\forall y\in K.
$$
Taking the supreme with respect to $y\in K$ we get
$$ f(x_0)\le f(x)+\epsilon. $$
The same argument shows that
$$ f(x)\le f(x_0)+\epsilon. $$
Putting it all together we get
$$
|x-x_0|<\delta\implies|f(x_0)-f(x)|\le\epsilon.
$$
