$\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\mc}{\mathcal}$
Question
Let $X$ be a compact Hausdorff space and $\Delta_2(X)$ denote the set $\set{(x, x):\ x\in X}$ in $X\times X=:X^2$.
I want to confirm that the following is true (a proof if supplied below).
Theorem 1.
Let $X$ be a compact Hausdorff space and $O$ be a neighborhood of $\Delta_2(X)$ in $X^2$.
Then there is a neighborhood $Q$ of $\Delta_2(X)$ such that whenever $(x, y)$ and $(y, z)$ are in $Q$ for some $x, y, z\in X$, we have $(x, z)\in O$.
The motivation for this is to generalize, to compact Hausdorff spaces, the following fact about metric spaces that if "$d(x, y), d(y, z)< \varepsilon/2$ then $d(x, z)< \varepsilon$."
My larger gaol was to have a device which allows mimicking proofs in topological dynamics for compact metric spaces to arbitrary compact Hausdorff spaces.
The purpose of this post is two-fold.
One is to verify my proof below, and the other is to get a shorter proof of the theorem above.
(If you do not want to read my proof and supply your own proof then please go ahead and share!)
I am somewhat apprehensive about my proof since it is longer that what seems necessary and also that it took me many iterations to get the details right, for I had incorrectly proven it multiple times in the process.
Purported Proof
Lemma 2.
Let $X$ be a compact Hausdorff space and $A$ be a closed set in $X$.
Let $U$ be neighborhood of $A$.
Then there is a neighborhood $O$ of $A$ in $X$ such that $\bar O\subseteq U$.
Proof.
Restatement of the fact that compact Hausdorff spaces are normal.
Lemma 3.
Let $X$ be a compact Hausdorff space and $O$ be a neighborhood of $\Delta_2(X)$ in $X^2$.
Then there is an open cover $\mc V$ of $X$ such that
$$
(V\cup V')\times (V\cup V') \subseteq O
$$
whenever $V, V'\in \mc V$ are such that $V\cap V'\neq \emptyset$.
Proof.
We say that an open cover $\mc U$ of $X$ if good if $\overline{\bigcup_{U\in \mc U} U\times U} $ is contained in $O$.
It is clear from Lemma 2 and from compactness of $X$ that finite good open covers of $X$ exist.
Also, given an open cover $\mc U$ of $X$, we say that $G$ in $\mc U$ is well-behaved if whenever $G\cap U\neq \emptyset$ for some $U$ in $\mc U$, we have $(G\cup U)\times (G\cup U)$ is contained in $O$.
Let $\mc U=\set{U_1, \ldots, U_m, G_1, \ldots, G_n}$ be a be an arbitrary finite good open cover of $X$, where each $G_i$ is well-behaved and each $U_i$ is not well-behaved.
If $m=0$ then we are done.
So assume that $m\geq 1$.
It automatically follows that then $m\geq 2$.
We will construct a finite good open cover of $X$ which has fewer ill-behaved elements.
By Lemma 2 we know that there is a neighborhood $Q$ of $\Delta_2(X)$ which contained $\overline{\bigcup_{U\in \mc U} U\times U}$ such that $\bar Q\subseteq O$.
Let $K$ be the boundary of $U_1\cup \cdots \cup U_{m-1}$.
For each $p$ in $K$, let $W_p$ be a neighborhood of $p$ in $X$ such that $W_p$ is contained in $G_i$ whenever $W_p\cap G_i\neq \emptyset$, and $(W_p\cup U_i)\times (W_p\cup U_i)\subseteq Q$ whenever $W_p\cap U_i$ is not empty.
The existence of $W_p$ can be established by a compactness argument.
Since $K$ is compact, there is a finite set $F$ of $K$ such that $\set{W_p:\ p\in F}$ covers $K$.
Define $U_m'=U_m\setminus \overline{U_1\cup \cdots \cup U_{m-1}}$ and
$$
\mc U'
=
\set{U_1, \ldots, U_{m-1}, U_m', G_1, \ldots, G_n} \cup \set{W_p:\ p\in F}
$$
It is easy to check that $U_m'$ as well as each $G_i$ is well-behaved in $\mc U'$.
Also, $\overline{\bigcup_{U'\in \mc U'} U'\times U'}$ is contained in $Q$, and hence $\mc U'$ is a good open cover.
This finishes the proof.
$\blacksquare$
Theorem 4.
Let $X$ be a compact Hausdorff space and $O$ be a neighborhood of $\Delta_2(X)$ in $X^2$.
Then there is a neighborhood $Q$ of $\Delta_2(X)$ such that whenever $(x, y)$ and $(y, z)$ are in $Q$ for some $x, y, z\in X$, we have $(x, z)\in O$.
Proof.
Let $\mc U$ be an open cover of $X$ such that whenever $U$ and $U'$ in $\mc U$ are such that $U\cap U' \neq \emptyset$, we have $(U\cup U')\times (U\cup U')$ is contained in $O$.
Such an open cover is furnished by Lemma 3.
Define $Q=\bigcup_{U\in \mc U} U\times U$.
Now let $(x, y)$ and $(y, z)$ be in $Q$ for some $x, y, z\in X$.
Then there are $U$ and $U'$ in $\mc U$ such that $(x, y)\in U\times U$ and $(y, z)\in U'\times U'$.
Thus $U\cap U'$ is non-empty, and thus $(U\times U')\times (U\times U')$ is contained in $O$.
But since $(x, z)$ is in $(U\cup U')\times (U\cup U')$, we see that $(x, z)\in O$, and we are done.
$\blacksquare$
Best Answer
In this old answer (almost a duplicate but not quite IMO) you'll find an alternative shorter proof. I said in the comments: this fact about diagonal neighbourhoods is part of a larger theory on uniformities, in particular the fact that compact Hausdorff spaces have a unique uniformity, i.e. the set of all diagonal neighbourhoods.