I'm trying the following problem: Let $\left(X, \tau \right)$ be a completely regular space and $A$ a closed subset of $X$. Show that $\left(X, \tau_A\right)$ is completely regular where $\tau_A=\{U\cup \left(V\cap A\right): U, V \in \tau \}$
So, I did the following: Let $F$ be a closed subset of $X$ with the new topology. So $X\setminus F$ is an open subset of $X$. Then, $X\setminus F= U\cup (V\cap A)$ for some $U$, $V$ in $\tau$. Let $x \in X\setminus F$. So we have two cases:
If $x\in U$ the result follows easy. The problem is when $x\in V\cap A$. I suspect that the fact of $A$ being closed is strong in this part but I can't see it. I tried with $X=\mathbb{R}$ and the usual topology and $[0,1]=A$. Notice that the set $K=(-\infty,0)\cup [\frac{1}{2},1]$ is closed in the new topology. In this point, is easy construct a continuous function such that $f(0)=0$ and $f(K)\subseteq \{1\}$. The new topology, as we can observe, solve some problems in the left limit: $f$ can be defined as the line segment between $(0,0)$ and $(\frac{1}{2}, 1)$ as points in the plane and $1$ in other case. This function is continuous. Is a interesting exercise, but I can't solve it in the second case.
Notation (from Set Theory): For function $h$ and $S\subset dom(h)$ write $h''S =\{h(s):s\in S\}.$ (Read as "$h$-double-prime-$S$".)
Let $p \in X .$ Let $C$ be $\tau_X$-closed with $p\not \in C.$
$(1).$ If $p\not \in A:$ Observe that a $\tau_A$-open-set that is disjoint from $A$ is a $\tau$-open-set, so $p\in (X \backslash A) \backslash C\in \tau,$ and there exists a $\tau$-continuous $f:X\to [0,1]$ with $f(p)=0$ and $\{1\}\supseteq f'' (X \backslash ((X \backslash A) \backslash C)=f'' (A \cup C)\supseteq f''C.$
Now $\tau_A\supset \tau$ so $f$ is also $\tau_A$-continuous.
$(2).$ If $p\in A:$ Take $V\in \tau$ such that $p\in V\cap A\subset X$ \ $C.$ Let $f:X\to [0,1]$ be $\tau$-continuous with $f(p)=0$ and $f''(X$ \ $V)\subseteq \{1\}. $
Now let $ g(q)=f(q)$ for $q\in A$ and let $g(q)=1$ for $q\in X$ \ $A .$
Prove that $g^{-1}[0,r)$ and $g^{-1}(r,1]$ belong to $\tau_A$ for every $r\in (0,1),$ so that $g$ is $\tau_A$-continuous. We have $g(p)=0$ and $g''C\subseteq \{1\}.$
(3). $\tau$ is a $T_1$ topology and $\tau\subset \tau_A$ so $\tau_A$ is also $T_1.$