Probably an important thing to investigate is the connection between the topologies $\tau$ and $\tau_A$.
- $\tau_A$ is finer than $\tau$. This follows from the fact that $U = U \cup ( \varnothing \cap A )$ for every $U \in \tau$.
- Since $\tau_A$ is finer than $\tau$, then $\operatorname{cl}_{\tau_A} ( B ) \subseteq \operatorname{cl}_{\tau} ( B )$ for every $B \subseteq X$ (where $\operatorname{cl}_\sigma ( B )$ denotes the closure of $B$ with respect to the topology $\sigma$).
Next, it is probably easier to prove this using the open-neighbourhood characterisation of regularity:
A T1-space $Y$ is regular if (and only if) given any $y \in Y$ and any open neighbourhood $U$ of $Y$ there is an open neighbourhood $W$ of $x$ such that $\operatorname{cl} (W) \subseteq U$.
We can now demonstrate that $\langle X , \tau_A \rangle$ is regular:
Let $x \in X$, and let $U \cup ( V \cap A )$ be a $\tau_A$-open neighbourhood of $x$ (where $U , V \in \tau$). There are two cases:
If $x \in U$, then by the regularity of $\langle X , \tau \rangle$ there is a $\tau$-open neighbourhood $W$ of $x$ such that $\operatorname{cl}_\tau ( W ) \subseteq U$. But then $W$ is also a $\tau_A$-open neighbourhood of $x$, and $\operatorname{cl}_{\tau_A} ( W ) \subseteq \operatorname{cl}_\tau ( W ) \subseteq U \subseteq U \cup ( V \cap A ).$
If $x \in V \cap A$, then $V$ is a $\tau$-open neighbourhood of $x$, so by the regularity of $\langle X , \tau \rangle$ there is a $\tau$-open neighbourhood $W$ of $x$ such that $\operatorname{cl}_\tau ( W ) \subseteq V$. Then $W \cap A$ is a $\tau_A$-open neighbourhood of $x$, and $ \operatorname{cl}_{\tau_A} ( W \cap A ) \subseteq \operatorname{cl}_\tau ( W \cap A ) \subseteq \operatorname{cl}_\tau ( W ) \cap \operatorname{cl}_\tau ( A ) \subseteq V \cap A \subseteq U \cup ( V \cap A ).$ (Though a bit hidden, we are using the fact that $A$ is closed, since then $\operatorname{cl}_\tau ( A ) = A$. Were $A$ not closed, the penultimate set-inclusion above may be false.)