I have some trouble finding a right way to solve this problem. The only thing I can think of is the Tietze Extension Theorem, but somehow it seems that the converse of Tietze Extension Theorem is more useful.
Suppose $X$ is Tychonoff $(T_1)$ and $X$ has the property that every continuous, bounded real valued function on a closed subset of $X$ has a continuous extension to all of $X$. We want to show that $X$ is normal.
My attempt: Since $X$ is Tychonoff, then for any $x,y\in X$ with $x\neq y$, there is a neighbourhood $U$ of $x$ such that $x\in U$ but $y\notin U$. Let $f$ be any continuous, bounded real valued function on a closed subset $Y$ of $X$, i.e. $f:Y\to \mathbb{R}$ and its continuous extension be $F:X\to\mathbb{R}$.
To show $X$ is normal, let $A$ be a closed subset of $X$ and $X\cap A=\emptyset$. I am not sure how can we continue from here.
How can we construct the neighbourhoods of $X$ and $A$? How can we make use of $f$ and $F$?
Could anyone please give some concrete hints? I have spent some time on this problem but still couldn't see any light. Thanks.
Let $A,B$ be two disjoint closed subsets of $X$.
check:$f:A\cup B\rightarrow \mathbb{R} $ with $f(A)=\{0\}$ and $f(B)=\{1\}$ is a continuous function.
So by assumption, you get the continuous extension $F:X\rightarrow \mathbb{R} $ with $F|_{A\cup B}=f$, then consider $F^{-1}((-\frac{1}{2},\frac{1}{2}))$ and $F^{-1}((\frac{1}{2},2))$, which are disjoint open sets separating $A$ and $B$.
That means $X$ is normal.