Normal Hausdorff implies functionally Hausdorff

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., normal space) must also satisfy the second topological space property (i.e., Urysohn space)
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Statement

Any normal space is a Urysohn space.

Definitions used

Normal space

Further information: Normal space

A normal space is a topological space $X$ that is a T1 space and satisfies the following: for any two disjoint closed subsets $A,B$ of $X$, there are disjoint open subsets $U,V$ of $X$ containing $A,B$ respectively.

Urysohn space

Further information: Urysohn space

A Urysohn space is a topological space $X$ such that, for any two distinct points $x,y\in X$, there is a continuous function $f:X\to [0,1]$ such that $f(x)=0$ and $f(y)=1$.

Facts used

1. Urysohn's lemma: If $X$ is a normal space and $A,B$ are disjoint closed subsets of $X$, then there exists a continuous function $f:X\to [0,1]$ such that $f(a)=0$ for all $a\in A$ and $f(b)=1$ for all $b\in B$.

Proof

Given: A topological space $X$ that is normal.

To prove: $X$ is a Urysohn space.

Proof: Suppose $x,y$ are distinct points of $X$. Since $X$ is normal, it is $T_{1}$, and $\{x\},\{y\}$ are both closed points. Fact (1) then provides the function $f:X\to [0,1]$ with $f(x)=0$ and $f(y)=1$.