Normal Hausdorff implies functionally Hausdorff

Statement
Any normal space is a Urysohn space.

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
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) uses::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 $$f(y) = 1$$.