order topology is Hausdorff Is it appropriate to ask the community to check my proof?  I am rereading Munkres Topology and trying to do the HW.  This is my attempt for #10 on page 101.

Show that every order topology is Hausdorff.

Proof:
Suppose that $x_1, x_2$ are elements of $X$, and $x_1 < x_2$.
Case 1:
Suppose that $x_2$ is the next element after $x_1$, and suppose that $$x_0 < x_1 < x_2 < x_3.$$  Then $x_1$ is an element of $(x_0, x_2) = U_1$.  Then $x_2$ is an element of $(x_1, x_3) = U_2$.
$U_1$ and $U_2$ are disjoint.
Case 2:
Suppose that there might not be "next elements" in $X$ but $x_1 < x_2$.  Then there exists $$a < x_1 < b < x_2 < c.$$  Then $x_1$ is an element of $(a, b) = U_1$.  Then $x_2$ is an element of $(b, c) = U_2$.
$U_1$ and $U_2$ are disjoint.
Thus $X$ is Hausdorff.
Is this on the right track?
 A: Looks almost correct to me.
It might be a bit tighter to phrase you first case not in terms of "next elements," but in terms of whether a there exists an element $b$ such that $x_1 < b < x_2$.
Also, you don't cover cases where $x_1$ is minimal or $x_2$ is maximal. Those should be easy to do from the definition of the order topology, though.
A: It’s almost right: in Case $1$ you forgot to cover the cases in which $x_1$ or $x_2$ is an endpoint of the space. However, there’s no need to treat these separately if you modify your proof a little. I’d arrange it like this:

Let $x,y\in X$, and assume without loss of generality that $x<y$. If there is a $z\in X$ such that $x<z<y$, let $U=(\leftarrow,z)$ and $V=(z,\to)$; then $x\in U$, $y\in V$, and $U\cap V=\varnothing$. If there is no such $z$, let $U=(\leftarrow,y)=(\leftarrow,x]$ and $V=(x,\to)=[y,\to)$; once again $x\in U$, $y\in V$, and $U\cap V=\varnothing$, and $X$ is therefore Hausdorff.

Here $(z,\to)=\{u\in X:u>z\}$, $(\leftarrow,z)=\{u\in X:u<z\}$, $[z,\rightarrow)=\{u\in X:u\ge z\}$, and so on.
