Let $X$ be a normal topological space. Tietze's extension theorem says that if $A \subset X$ is closed, then a continuous function $f: A \to \mathbb R^n$ can be extended to a continuous function whose domain is all of $X$. If $X$ is assumed to be a metric space, then the theorem holds for functions taking values in any locally convex linear space (see this).
I am wondering if the theorem holds for certain set-valued functions.
In particular, let $\phi$ be a function from closed $A \subset X$ into compact subsets of $\mathbb R^n$. To say that $\phi$ is continuous on $A$ means that the following conditions are met for every $x \in A$:
(1) For every neighborhood $U$ of $\phi(x)$, there is a neighborhood $V$ of $x$ such that $\phi(y) \subset U$ for all $y \in V$;
(2) For every open subset $U$ of $\mathbb R^n$ for which $\phi(x) \cap U \neq \emptyset$, there is a neighborhood of $V$ of $x$ such that $\phi(y) \cap U \neq \emptyset$ for all $y \in V$.
Can $\phi$ be extended to a continuous set-valued function whose domain is all of $X$?
In principle, I don't mind assuming that $X$ is actually a subset of $\mathbb R^m$.