Normal Space is Tychonoff Space

Theorem
Let $\left({S, \tau}\right)$ be a normal space.

Then $\left({S, \tau}\right)$ is also a Tychonoff (completely regular) space.

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

From the definition of normal space:
 * $\left({S, \tau}\right)$ is a $T_4$ space
 * $\left({S, \tau}\right)$ is a $T_1$ (Fréchet) space.

Let $F$ be any closed set in $T$.

Let $y \in \complement_S \left({F}\right)$, that is, $y \in S$ such that $y \notin F$.

As $T$ is a $T_1$ (Fréchet) space it follows from Equivalence of Definitions of $T_1$ Space that $\left\{{y}\right\}$ is closed.

As $T = \left({S, \tau}\right)$ is a $T_4$ space, we have that for any two disjoint closed sets $A, B \subseteq S$ there exists an Urysohn function for $A$ and $B$.

But $F$ and $\left\{{y}\right\}$ are disjoint closed sets.

So there exists an Urysohn function for $F$ and $\left\{{y}\right\}$.

This is the definition of a $T_{3 \frac 1 2}$ space.

Next we note that as $T$ is a $T_1$ (Fréchet) space, from $T_1$ Space is $T_0$ Space it follows that $T$ is a $T_0$ (Kolmogorov) space.

So $T$ is both a $T_{3 \frac 1 2}$ space and $T_0$ (Kolmogorov) space.

So, by definition, $T$ is a Tychonoff space.