There’s a bit of sloppiness right off the bat in the definition of $\mathscr{C}_X$ as the set of all compactifications of $X$: they really mean that $\mathscr{C}_X$ is a set containing one representative of each homeomorphism class of compactifications of $X$. That is, each member of $\mathscr{C}_X$ is a compactification of $X$, and every compactification of $X$ is homeomorphic to exactly one member of $\mathscr{C}_X$. It actually takes a bit of set-theoretic finesse to justify the existence of this set, but never mind; we’ll take all that as read and worry about the topology.
Let $Y=\prod\mathscr{C}_X$, and let $\varphi:X\to Y:x\mapsto\big\langle c(x):cX\in\mathscr{C}_X\big\rangle$. For each $cX\in\mathscr{C}_X$, $c:X\to cX$ is an embedding, so it’s clear that $\varphi$ is an injection. To show that $\varphi$ is continuous, it suffices to show that for each $cX\in\mathscr{C}_X$ and each open set $U$ in $cX$, $\varphi^{-1}\left[\{y\in Y:y_{cX}\in U\}\right]$ is open in $X$, since sets of the form $\{y\in Y:y_{cX}\in U\}$ are a subbase for the product topology on $Y$. But
$\begin{align*} \varphi^{-1}\left[\{y\in Y:y_{cX}\in U\}\right]&=\{x\in X:c(x)\in U\}\\ &=c^{-1}[U]\\ &=c[X]\cap U\;, \end{align*}$
which is open in $X$ simply because $cX$ is a compactification of $X$.
(Here $c[X]=\{c(x):x\in X\}\subseteq cX$.)
To show that $\varphi$ is also open, suppose that $U$ is an open set in $X$. then
$\begin{align*} \varphi[U]&=\{\varphi(x):x\in U\}\\ &=\{\langle c(x):cX\in\mathscr{C}_X\rangle:x\in U\}\\ &=\{\langle c(x):cX\in\mathscr{C}_X\rangle:c(x)\in cU\}\;. \end{align*}$
If $cX,c'X\in\mathscr{C}_X$ and $x\in X$, $c(x)\in cU$ iff $c'(x)\in c'U$, so $\varphi[U]=\varphi[X]\cap\{y\in Y:y_{cX}\in cU\}$, where $cX$ is a single, fixed compactification of $X$. And $\{y\in Y:y_{cX}\in cU\}$ is a basic open set in the product $Y$, so $\varphi[U]$ is open in $\varphi[X]$. Thus, $\varphi$ maps $x$ homeomorphically to $\varphi[X]$.
Since $Y$ is compact, $\operatorname{cl}_Y\varphi[X]$ is certainly a compactification of $X$; call it $K$. All that’s needed to finish the argument is to show that for each $cX\in\mathscr{C}_X$ there is a continuous surjection $f:K\to cX$ that fixes $X$ pointwise. (Actually, that last bit is verbal sloppiness: what’s really meant is that for each $x\in X$, $f(\varphi(x))=c(x)$.) We can simply take $f$ to be the projection map from $Y$ to the factor $cX$, restricted to the subspace $K$: $f=\pi_{cX}\upharpoonright K$. Projections are always continuous, and $f(\varphi(x))=c(x)$ by the definition of $\varphi$, so we’re done: $K$ is the Čech-Stone compactification of $X$.