Let $X$ be a variety such that $\forall x,y\in X$ there exists an open affine $U\in X$ containing $x,y$. Then $X$ is separated I am struggling to prove this. Note that I do not know anything about schemes, so please no schemes. 
I know that in order to show $X$ is separated I need to show that $\Delta_{X}=\{(x,x)\in X\times X\}$ is closed in $X\times X$. I'm really stuck...
A hint or answer would be greatly appreciated.

You want to show that the complement of $\Delta_X$ is open in $X\times X$. Let $x\ne y$, then $(x,y)$ is in the complement of $\Delta_X$, so we want to produce an open neighborhood of $(x,y)$ not intersecting $\Delta_X$. If $x,y\in U$, then it suffices to show that $(U\times U) - \Delta_X = (U\times U) - \Delta_U$ is open, but this is automatic, since affine varieties are always separated.
