Show a Subspace of regular space is regular 
I use this definition .Proofs of this theorem
I saw are  based on a similar def,  but with functions.
Source:
(https://www.math.tamu.edu/~tomzz/math636/ass7.pdf)
Attempt
Let (X$,\mathcal{T}$) be a regular topology.
Let $\mathcal{T}_S$={O$\cap$ S :O$\in\tau$}
Suppose X is a regular and $S\subseteq X$,S closed and any point  x$\in A$ and V$\subset A$
Let x$\in S\setminus A$, then $S\setminus A$ is open
Then there exists U open set in X s.t U $\cap S$ =V
Clearly $ x\notin$ U else x$\in V$
Since X is regular  (S$\setminus A )\cap S=\emptyset$
 A: I'm not following your proof.  You need to show that a subspace of X is regular.   You need to give it a name, if you are using $A$ that might confuse you with the $A$ in the theorem so I suggest calling it something else, like let $S\subseteq T$ be equipped with the subspace topology (IE a set is open in $S$ if and only if it is equal to the intersection of an open set $U$ in $T$ with $S$).
Now, within $S$ you need to show regularity, so you need to take a closed set in $S$, call it $A\subseteq S$ and a point  $x\in S\setminus A$  and show there exists open sets $U,V$ in the subspace topology that separate $A$ and $x$ in $S$.  Hint: Consider what it means to be closed set in $S$ compared to $T$,  then use the regularity of $T$ to generate $U',V'$ in $T$ that you can then derive your $U,V$ in S from.
A: Do the proof more systematically. Don’t needlessly  repeat definitions.
Let $(X,\tau)$ be a regular space and let $S \subseteq X$ be a subset in the subspace topology.
Let $x \in S$ and let $C \subseteq S$ be closed in $S$ such that $x \notin C$.
By standard facts about the subspace topology, there is a closed subset $C’$ of $X$ such that $$C = C’ \cap S$$
It’s clear that $x \notin C’$ as well, so by regularity of $X$ there are open sets $U$ and $V$ of $X$ such that $x \in U$, $C’ \subseteq V$ and $U \cap V = \emptyset$.
Then $U’ = U \cap S$ is open in $S$ and contains $x$, $V’=V \cap S$ is open in $S$ and $$C = C’ \cap S \subseteq V \cap S = V’$$
Of course $U’ \cap V’ \subseteq U \cap V = \emptyset$ so these sets show that $S$ is regular.
