I consider $(X,d_{X})$ and $(Y,d_{Y})$ two metric spaces, the first being compact and $T : X \rightrightarrows Y$ a multi valued function which is upper semi continuous and with compact values (i.e $T(x)$ is compact).
I would like to prove that $T(X)=\cup_{x\in X}T(x)$ is a compact set.
Here is my attempt :
I will use sequential compactness. Consider $y_n$ a sequence in $T(X)$. We associate (use of the choice axiom) to each $y_n$ an $x_n$. This gives us a sequence $x_n$ in $X$. By compactness of $X$ we extract a subsequence $x_{\phi(n)}$ that converges to $x$.
From there, my argument is sloppy : I would like to use the upper semi continuity that is that for all open set $U$ in $Y$ the set $\{x : T(x)\subset U\}$ is open in $X$.
However I cannot see how to use this.
The idea I have in mind is that after some rank, $T(x_{\phi(n)})\in T(\bar{x})$ then the conclusion would be easy, but this configuration seems unlikely.
Here is one way to argue. Since $Y$ is metrizable, it suffices to check sequential compactness of $T(X)$. Consider a sequence $y_n\in T(X)$, $y_n\in T(x_n)$ for some $x_n\in X$. By compactness of $X$, after extraction, we can assume that $x_n\to x\in X$. Take $U_n\subset Y$, the $1/n$-neighborhoods of $T(x)$. By the upper semicontinuity of $T$, for each $n$ there exists $N=N(n)$ such that $y_k\in T(x_k)\in U_n$ for all $k\ge N$. Let $z_k\in T(x)$ be a point nearest to $y_k$. Thus, $d(z_N,y_N)\le 1/n$. By compactness of $T(x)$, after extraction, we have $z_N\to z\in T(x)$. Hence, by the triangle inequality, $y_N\to z$ as well. Thus, $T(X)$ is compact. qed