$f(x) = \inf_{y \in Y} c(x,y) - \inf_{\xi \in X} c(\xi,y) - f(\xi) \Rightarrow f$ is upper semicontinuous Let $X, Y$ be metric spaces. Given $c: X \times Y \mapsto \mathbb{R}$ continuous, define
$$ f(x) = \inf_{y \in Y} \left( c(x,y) - \inf_{\xi \in X} (c(\xi,y) - f(\xi)) \right).$$
Then is $f$ upper semi-continuous? Why?
 A: Recall that $f:\mathbb{R}\longrightarrow [-\infty,+\infty]$ is upper semi continuous if and only if $f^{-1}(-\infty,\alpha)$ is open for every $\alpha\in\mathbb{R}$. If you did not know that, it is not hard to prove from any definition of upper semicontinuity you might have. And that's should be your first step, as it is the most tractable characterization of uppersemicontinuity.
Claim: If $(f_i)_{i\in I}$ is a family of upper semi continuoous functions, then $f:=\inf_i f_i$ is upper semi continuous.
Proof: It is easily seen that $\inf_i f_i(x)<\alpha$ if and only if there exists $i\in I$ such that $f_i(x)<\alpha$. In other terms
$$
f^{-1}(-\infty,\alpha)=\bigcup_{i\in I}f_i^{-1}(-\infty,\alpha).
$$
By assumption, every set in this union is open, so $f^{-1}(-\infty,\alpha)$ is open for every $\alpha\in\mathbb{R}$. That is $f$ is upper semi continuous. QED.
Note that a continuous function is upper semi continuous. Now apply this claim to the continuous functions
$$
f_y(x):=c(x,y)-\inf_\xi (c(\xi,y)-f(\xi))=c(x,y)- K_y
$$
where $K_y$ is a constant independent of $x$. You get that $f=\inf_y f_y$ is upper semi continuous.
