Urysohn Space is Completely Hausdorff Space

Theorem
Let $\left({X, \vartheta}\right)$ be an Urysohn space.

Then $\left({X, \vartheta}\right)$ is also an $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Proof
Let $T = \left({X, \vartheta}\right)$ be an Urysohn space.

Then for any distinct points $x, y \in X$ (i.e. $x \ne y$), there exists an Urysohn function for $\left\{{x}\right\}$ and $\left\{{y}\right\}$.

Thus:
 * $\forall x, y \in X: x \ne y: \exists U, V \in \vartheta: x \in U, y \in V: U^- \cap V^- = \varnothing$

which is precisely the definition of an $T_{2 \frac 1 2}$ (completely Hausdorff) space.