Let X be a $T_1$-space satisfying the conclusion of Tietze's Extension Theorem then prove that X is normal I am unsure what exactly the "conclusion" of the theorem means, as in, is it the entire theorem without the assumption of normality of X? So I am assuming that we have to go the other way around as compared to the proof of Tietze's theorem, i.e, assume the function f on (closed?) subspace F has a continuous extension defined on all of X (both of who's values lie in [a,b]) and go backward to prove X is normal. Also, in particular it mentions $T_1$-space, why not a general topological space?
The entire proof of Tietze's Theorem is based on the assumption of normality of X, so I cannot understand how to go about proving the reverse to arrive at the condition of normality.

Let $A$ and $B$ be disjoint closed sets. Define $f:A \cup B \to \mathbb R$ by $f(x)=0$ if $x \in A$ and $f(x)=1$ if $x\in B$. Then we can verify that $f$ is continuous by showing that the inverse image of any closed set is closed. We are told that Tietze's Theorem can applied. This means any continuous function defined on  a closed subset of $X$ can be extended to a continuous function on $X$. So there is a continuous function $F: X \to \mathbb R$ such that $F(x)=0$ if $x \in A$ and $F(x)=1$ if $x\in B$. Now $F^{-1} (-\infty, \frac 1 2)$ and $F^{-1}(\frac 12 , \infty)$ are disjoint open sets containing $A$ and $B$ respectively. This proves that $X$ is normal. 
