Completely Hausdorff Space is Preserved under Closed Bijection

Theorem
Let $T_A = \left({X_A, \vartheta_A}\right), T_B = \left({X_B, \vartheta_B}\right)$ be topological spaces.

Let $\phi: T_A \to T_B$ be a closed bijection.

If $T_A$ is a $T_{2 \frac 1 2}$ (Urysohn) space, then so is $T_B$.

Proof
Let $T_A$ be a $T_{2 \frac 1 2}$ (Urysohn) space.

Then:
 * $\forall x, y \in X_A, x \ne y: \exists U_A, V_A \in \vartheta_A: x \in U_A, y \in V_A: U_A^- \cap V_A^- = \varnothing$