Let $(X,d)$ be a complete metric space, $CB(X)$ the set of closed and bounded subsets of $X$, and $T:X\rightarrow C(X)$ be a multi-valued function.
How can you prove this:
If $T$ is upper semi-continuous then $f(x)=d(x,T(x))$ is lower semi-continuous
And the 2nd question is that Does the reverse hold?
I hope that the definition of "upper semi-continuous" is the following $T:X\to CB(X)$ is usc if $\{ x;\; T(x)\subset U\}$ is open for every open set $U\subset X$.
Assume that $T:X\to CB(X)$ is usc. We have to show that for any $\alpha\in\mathbb R$, the set $O_\alpha=\{ x;\, d(x,T(x))>\alpha\}$ is open in $X$. Fix $\alpha$ and take any $x_0\in O_\alpha$. Choose $\varepsilon >0$ such that $d(x_0,T(x_0))>\alpha+\varepsilon$. Then $T(x_0)\subset U_0=X\setminus B_0$, where $B_0$ is the closed ball $B(x_0,\alpha+\varepsilon]$. Since $U_0$ is an open set, the set $V_0=\{ x;\; T(x)\subset U_0\}$ is an open neighbourhood of $x_0$; and by definition we have $d(x_0, T(x))\geq \alpha+\varepsilon$ for every $x\in V_0$. By the triangle inequality follows that if $x\in V_0$ and $d(x,x_0)<\varepsilon$, then $d(x, T(x))> \alpha$, i.e. $x\in O_\alpha$. Hence, $V_0\cap B(x_0,\varepsilon)$ is an open neighbourhood of $x_0$ contained in $O_\alpha$. This shows that $O_\alpha$ is open in $X$, for any $\alpha$.
The converse is not true. Take for example $X=\mathbb R$. Let $f:\mathbb R\to\mathbb R$ be any lower semi-continuous but not continuous map such that $f(x)>x$ for all $x\in X$, and define $T(x)=\{ f(x)\}$. Then $T$ is not usc, but $d(x,T(x))=\vert x-f(x)\vert=f(x)-x$ is sci.