Urysohn is refining-preserved

Statement
If $$X$$ is a Urysohn space with a topology $$\tau$$, and if $$\tau'$$ is a finer topology than $$\tau$$, then $$X$$ is a Urysohn space with topology $$\tau'$$.

Related facts

 * Hausdorffness is refining-preserved
 * Regularity is not refining-preserved
 * Complete regularity is not refining-preserved

Proof
Given: A topological space $$(X,\tau)$$. $$\tau'$$ is a finer topology than $$\tau$$. $$X$$ is a Urysohn space with topology $$\tau$$.

To prove: $$(X,\tau')$$ is a Urysohn space: for distinct points $$x,y \in X$$, there exists a function $$f':X \to [0,1]$$ that is continuous with respect to $$\tau$$ such that $$\! f'(x) = 0$$ and $$\! f'(y) = 1$$.

Proof: We have a continuous function $$\! f:(X,\tau) \to [0,1]$$ such that $$\! f(x) = 0$$ and $$\! f(y) = 1$$, continuous with topology $$\tau$$. Since $$\tau'$$ is finer than $$\tau$$, the identity map $$(X,\tau') \to (X,\tau)$$ is continuous. Composing with $$f$$, we obtain a map $$\! f':X \to [0,1]$$ such that $$\! f(x) = 0$$ and $$\! f(y) = 1$$.