Prove that if $X$ is Hausdorff, $\Delta$ is closed in $X\times X$ Prove that if $X$ is Hausdorff, $\Delta=\{(x, x)\mid x\in X\}$ is closed in $X\times X$ (with the product topology).
My attempt:
Let $x_1, x_2\in X$ s.t. $x_1\ne x_2$.
There exist neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$ that are disjoint.
$U_1\times U_2$ is a basis element in the product topology on $X\times X$. So, $U_1\times U_2$ is open in $X\times X$.
Let $x\in X$. 
$(x, x)\in U_1\times U_2\implies x\in U_1$ and $x\in U_2\implies x\in U_1\cap U_2$, which contradicts the fact that $U_1$ and $U_2$ are disjoint.
So, $(x, x)\notin U_1\times U_2$.
I feel that I'm on the right track but don't know how to proceed. Could someone please help me out?
 A: You're basically there, you just need to interpret your result. You found that for any point $(x_1,x_2)\in X\times X-\Delta$, there exists a neighborhood of $(x_1,x_2)$ contained in $X\times X-\Delta$. That is, $X\times X-\Delta$ is open. Therefore $\Delta$ is...
A: Your work shows that
$$
\Delta^c=\bigcup_{\substack{(x_1,x_2)\in X\times X\\x_1\neq x_2}} U_1(x_1)\times U_2(x_2),
$$
where $U_1(x_1)$ and $U_2(x_2)$ are separating sets for $x_1,x_2$. This shows the complement of $\Delta$ is a union of open sets, so the complement of $\Delta$ is open, so $\Delta$ is closed.
A: Let $(x, y)\in X\times X-\Delta$
$\implies(x, y)\in X\times X$ and $(x, y)\notin\Delta$
$\implies x, y\in X\text{ and }x\ne y$
There exist neighborhoods $U_x$ and $U_y$ of $x$ and $y$ respectively that are disjoint.
$U_x\times U_y$ is a basis element in the product topology on $X\times X$.
$(x, y)\in U_x\times U_y$ --------------------------------------------- (1)
Let $(u, v)\in U_x\times U_y$. --------------------------------------------- (2)
$\implies u\in U_x$ and $v\in U_y$
Since $U_x\subset X$ and $U_y\subset X$,
$u, v\in X$
$\implies(u, v)\in X\times X$ --------------------------------------------- (3)
$u=v\implies u=v\in U_x\cap U_y$, a contradiction as $U_x$ and $U_y$ are disjoint.
So, $u\ne v$.
$\implies(u, v)\notin\Delta$ --------------------------------------------- (4)
From (3) and (4),
$(u, v)\in X\times X-\Delta$ --------------------------------------------- (5)
From (2) and (5),
$U_x\times U_y\subset X\times X-\Delta$ --------------------------------------------- (6)
Combining (1) and (6),
$(x, y)\in U_x\times U_y\subset X\times X-\Delta$
So, $X\times X-\Delta$ is open in $X\times X$.
So, $\Delta$ is closed in $X\times X$.
QED
