Urysohn's Lemma Converse

Lemma
Let $T = \left({S, \tau}\right)$ be a topological space.

Let there exist an Urysohn function for any two $A, B \subseteq S$ which are closed sets in $T$ such that $A \cap B = \varnothing$.

Then $T = \left({S, \tau}\right)$ is a $T_4$ space.

Proof
Let $A$ and $B$ be arbitrary closed sets of $T$ and let $f$ be an Urysohn function for $A$ and $B$.

Let $C = \left[{0 \,.\,.\,\dfrac 1 4}\right)$ and $D = \left({\dfrac 3 4 \,.\,.\, 1}\right]$.

Then $C$ and $D$ are open in $\left[{0 \,.\,.\, 1}\right]$.

Hence, $f^{-1} \left({C}\right)$ and $f^{-1} \left({D}\right)$ are open in $T$.

Furthermore, by definition of Urysohn function, $A \subset f^{-1} \left({C}\right)$ and $B \subset f^{-1} \left({D}\right)$.

Also, from Preimage of Intersection under Mapping:
 * $C \cap D = \varnothing \implies f^{-1} \left({C}\right) \cap f^{-1}\left({D}\right) = \varnothing$

Therefore, $T$ is a $T_4$ space.

Also see

 * Urysohn's Lemma