Prove that every simply ordered set is a Hausdorff space in the order topology. The product of two Hausdorff space is again a Hausdorff space

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In the book of General Topology by Munkres, at page 100, it is asked to prove that

Every simply ordered set is a Hausdorff space in the order topology.The product of two Hausdorff space is again a Hausdorff space.

Proof of the first statement:

Let X be a topological space with the order topology and $a,b \in X$. Without loss of generality, assume $a < b$, then since $(-\infty, b)$ is a $a$ neighbourhood and $(a,\infty)$ is a $b$ neighbourhood, and the intersection of these intervals are empty, we have the desired result.

For the second statement:

Let $X ,Y$ be Hausdorff spaces, then consider $(a,b), (c,d) \in X\times Y$, then by our hypothesis, $\exists U_a, V_b, U_b, V_d$ such that $U_a \cap U_c = \emptyset$ and $V_b \cap V_d = \emptyset$. Then $U_a \times V_b \cap U_c \times V_d = \emptyset$, hence $X\times Y$ is Hausdorff.


So, is there any problem, flow in the proof ? or any point that you advise me to clarify?

I mean even though the proofs are not complex, I have started Hausdorff spaces just today, and I want to make sure that that I'm not missing anything.