If every point of a topological space $X$ has a closed neighbourhood that is $T_2$ (as subspace), then $X$ is $T_2$. I have tried the problem in the following way-
Let $x,y$ be two distinct points in $X$ and let $x\in A$ and $y\in B$ where $A,B$ are closed neighbourhoods of $x,y$ respectively and $T_2$. Let $U_1$ and $U_2$ be open sets such that $x\in U_1\subset A$ and $y\in U_2\subset B$.
Case 1: If $y\notin A$, then take $V=A^c$ and $U=U_1$. Then $U\cap V=\emptyset $ and $y\in V$ and $x\in U$.
Case 2: If $x\notin B$ then take $U=B^c$, $V=V_1$. Then $U\cap V=\emptyset$ and $x\in U$, $y\in V$.
But I'm stuck with the case $x\in B$ and $y\in A$ i.e. $x,y\in A\cap B$.
Can anyone help me in this regard? Thanks for your help in advance.

In that last case: $A \cap B$ is Hausdorff, so there are open (in $A \cap B$) sets $O_1, O_2$ containing $x$ resp. $y$ that are disjoint.
So $O_i = (A \cap B ) \cap O'_i$, $O'_i$ open in $X$, $i=1,2$.
But then $U_1 \cap O'_1$ and $U_2 \cap O'_2$ disjoint open neighbourhoods of $x,y$ are as required.
