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.