My friend was reading a proof from "An Introduction to Dynamical Systems" from Michael Brin, and he got stuck on a part of the proof that boiled down to the following:
Let $X$ be a compact hausdorff topologial space. We'll denote by $\Delta$ the diagonal of $X$, i.e, $\Delta \doteq \{(x, x) \in X \times X \ \vert \ x \in X \}. $ Let $x \in X$ and let $U \ni x$ be a fixed neighbourhood of $x$.
Lemma. There is a neighbourhood of $x$, which we'll denote by $U'$, and there's an open set $V$ of $X \times X$, which satisfy the following:
i. $U' \subset U$ and $\Delta \subset V$
$x_1 \in U'$ and $(x_1, x_2) \in V \implies x_2 \in U$
For anyone interested, this is the part of the proof I mentioned:
We tried (Rodrigo tried, actually...) using the tube lemma but that led nowhere. The greatest difficulty we're facing is finding a useful neighbourhood of the diagonal, in fact we are not seeing any non trivial at all.
Just let $U'$ be any open neighborhood of $x$ whose closure is contained in $U$ and let $V=X\times X\setminus \left(\overline{U'}\times (X\setminus U)\right)$. Then $V$ is an open set containing $\Delta$, and if $x_1\in U'$ and $(x_1,x_2)\in V$, we must have $x_2\in U$ or else $(x_1,x_2)$ would be in $\overline{U'}\times(X\setminus U)$.
(So, this works in any regular space, since that's all you need in order to get such a $U'$.)