Let $F : X → X$ be continuous. Prove that the set $\{x ∈ X : F(x) = x\}$ of fixed points of F is closed in X Here X is a Hausdorff Space.
I know that singleton sets, {x}, are closed in a Hausdorff space. Although Im not sure if thats how to use the Hausdorff property. 
Should I investigate $h=F(x)-x$?
Can anyone give a hint
 A: In general you can’t investigate $F(x)-x$, since the operation of subtraction may not make any sense in the space $X$.
HINT: One straightforward approach is to show that the set of non-fixed points of $F$ is open. Suppose that $F(x)\ne x$. $X$ is Hausdorff, so there are open sets $U$ and $V$ such that $x\in U$, $F(x)\in V$, and $U\cap V=\varnothing$. Use these open sets and the continuity of $F$ to show that there is an open nbhd $W$ of $x$ such that $F(y)\ne y$ for each $y\in W$.
If you get completely stuck, there’s a further hint in the spoiler-protected block below.

 Show that you can find $W$ such that $x\in W\subseteq U$ and $F[W]\subseteq V$.

A: A space $X$ is Hausdorff iff the diagonal $\Delta$ is closed in $X \times X$.
Now the set of fixed points is exactly $(F\times Id_X)^{-1}(\Delta)$ which is closed by continuity of $F$ (and thus that of $F\times Id_X$).
A: I have this idea that I'd love somebody with more knowledge than I will check: define a function 
$$\;\phi:X\to X\times X\;,\;\;\text{by}\;\;\;\phi(x):=(x,F(x))\;$$
Since each coordinate function is continous also $\;\phi\;$ is, and if $\;\Delta:=\{(x,x)\in X\times X\}\;$ is the diagonal in the cartesian product, then
$$\phi^{-1}(\Delta\cap\phi(X))=\left\{x\in X\;;\;F(x)=x\;\right\}=:K$$
Since $\;\Delta\cap\phi(X)\;$ is closed  (see comments by Ayman Hourieh and Thomas Andres below the question) and $\;\phi\;$ is continous, then $\;K\;$ is closed.
