If for each $p\in X$, $\cap\{\overline{U}\mid U\in\Gamma,p\in U\}=\{p\}$, then the diagonal is closed I have the following problem,

Let $(X,\Gamma)$ is a topological space, if for each $p\in X$, $\cap\{\overline{U}\mid U\in\Gamma,p\in U\}=\{p\}$, then the diagonal $\{(x,x)\mid x\in X\}$ is closed in $X\times X$

My idea was, show that the condition "for each $p\in X$, $\cap\{\overline{U}\mid U\in\Gamma,p\in U\}=\{p\}$" is equivalent to say that $(X,\Gamma)$ is a hausdorff space, because the second condition (closed diagonal), is right if $(X,\Gamma)$ is a hausdorff space. This correct? or maybe exist straightforward way of prove this, Thanks! 
 A: We will prove that $\Delta(X)$ is closed by showing that it contains all its limit points.  Let $(a,b)$ be a limit point of the diagonal $\Delta(X)$.  We must show that $(a,b) \in \Delta(X)$, that is, that $a=b$.  Suppose not, so that $a \neq b$.
Claim: $b \in \bigcap \{\bar{U} : U \text{ open about } a\}$.
(This will be a contradiction since this was supposed to contain $a$ alone, but $b$ is different from $a$.)
Proof of Claim:  We must show that $b \in \bar{U}$ for any open set $U$ about $a$.  Take such an open set $U$.  If $b \in \bar{U}$ we are done.  If $b \notin \bar{U}$ then $U \times (X - \bar{U})$ is an open set in $X \times X$ about $(a,b)$.  Since $(a,b)$ is a limit point of the diagonal, this open set must intersect the diagonal.  Hence there exists a point of the form $(w,w)$ in $U \times (X - \bar{U})$.  This says that $w \in U$ but $w \notin \bar{U}$, which is bogus since $U \subseteq \bar{U}$.  This proves the claim and finishes the problem.
I have no doubt that someone can write a slicker argument that avoids all the contradictions, getting something more direct and more satisfying.  But, it is late here.
A: Your idea is great. Let $p\neq q$ in $X.$ Then
$$ \bigcap \{ \overline{U} \text{ | } U \in \Gamma, p \in U\}=\{p\} \neq \{q\} =\bigcap \{ \overline{V} \text{ | } V \in \Gamma, q \in V\}.$$
Thus, there exist $U \in \Gamma$ such that $p \in U$ and $q \not\in \overline{U},$ and also $V \in \Gamma$ such that $q \in V$ and $p \not\in \overline{V}.$ Then $U \cap (X-\overline{V})$ is a neighborhood of $p$ and $V \cap (X-\overline{U})$ is a neighborhood of $q,$ and
$$ U \cap (X-\overline{U}) \cap V \cap (X-\overline{U}) \subset U \cap (X-U) \cap V \cap (X-V)=\emptyset.$$
Since $p\neq q$ in $X$ were arbitrary, then $X$ is Hausdorff and hence the diagonal $\Delta$ is closed in $X \times X.$
