Corollary of Tietze extension theorem The Tietze extension theorem states that a space $X$ is normal iff every continuous function $f:A \rightarrow \mathbb{R}$, with $A$ a closed subset of $X$, can be extended to a continuous function $g:X \rightarrow \mathbb{R}$. 
The proof can be found here. I came across the following proposition:

A space $X$ is normal iff every lower semi-continuous multi-valued map $F: X \rightarrow 2^\mathbb{R}$ with compact and convex images admits a continuous selection.

The statement maybe requires some additional definitions:


*

*A multi-valued map $F: X \rightarrow 2^Y$ is called lower semi-continuous if for every $G \subseteq Y$ open, $\left\{x \in X \mid F(x) \cap G \neq \emptyset\right\}$ is also an open set.

*If $F: X \rightarrow 2^Y$ is a lower semi-continuous multi-valued map, we call $f: X \rightarrow Y$ a continuous selection if $f$ is continuous and $f(x) \in F(x)$ holds for all $x \in X$.


I don't know if this is needed to prove the proposition, but I'll include it anyways; I've proved that $c(F)$, defined by $c(F)(x) = c(F(x))$ (where $c(A)$ denotes the convex hull of a set $A$), is lower semi-continuous if $F: X \rightarrow 2^Y$.
Any help with proving the blockquoted statement is appreciated, I don't really know how to get started. It does state however that the proof uses the Tietze extension theorem.
 A: The implication from right to left isn’t too hard.
Suppose that $X$ has the continuous selection property. Let $F$ be an arbitrary closed subset of $X$ and $f:F\to\Bbb R$ a continuous function. Let $h:\Bbb R\to(0,1)$ be a homeomorphism, and let $g=h\circ f$. Define
$$G:X\to\wp(\Bbb R):x\mapsto\begin{cases}
\{g(x)\},&\text{if }x\in F\\
[0,1],&\text{otherwise}\;.
\end{cases}$$
If $U\subseteq\Bbb R$ is open, 
$$\{x\in X:G(x)\cap U\ne\varnothing\}=\begin{cases}
g^{-1}[U]\cup(X\setminus F),&\text{if }U\cap[0,1]\ne\varnothing\\
\varnothing,&\text{otherwise}\;,
\end{cases}$$
so $G$ is lower semicontinuous and takes compact, convex values. Let $\sigma:X\to\Bbb R$ be a continuous selection for $G$; then $\sigma(x)$ is a continuous extension of $g$ to all of $X$, and $h^{-1}\circ\sigma$ is a continuous extension of $f$. It now follows from the Tietze extension theorem that $X$ is normal.
I don’t immediately see how to prove the opposite implication; I’ll give it some thought after I get some sleep.
A: I think that, following the same steps with the proof (b) $\to$ (a) of Theorem 3.2'' in Continuous selections I, you can show that for every finite open cover of $X$ there is a partition of unity subordinated to it. The latter implies that $X$ is normal.
Otherwise, it's much easier to prove that $X$ is normal iff for every separable complete metric space $Y$ and for every compact-valued lower semicontinuous $\Phi:X\to 2^Y$ there is a compact-valued upper semicontinuous selection for $\Phi$.
If you are interested in similar characterizations of normality, you can find in "A unified approach to continuous, measurable selections and selections for hyperspaces" (Theorem 6.5) a proof of the following result:
$X$ is normal with $dim(X)=0$ (covering dimension zero) iff for every separable complete metric space $Y$ and every compact-valued lower semicontinuous $\Phi:X\to 2^Y$ there is a single-valued continuous selection for $\Phi$.
I hope I was helpful.
A: This a selection theorem due to E. Michael (Theorem 3.1', Continuous selections I).
