Let $X$ be a space; suppose that $h : X \to Z$ is an imbedding of $X$ in the compact Hausdorff space $Z$. Then there exists a corresponding compactification $Y$ of $X$; it has the property that there is an imbedding $H : Y \to Z$ that equals $h$ on $X$. The compactification $Y$ is uniquely determined up to an equivalence.
Given $h$, let $X_0$ denote the subspace $h(X)$ of $Z$, and let $Y_0$ denote its closure $Z$. Then $Y_0$ is a compact Hausdorff space and $\overline{X}_0$; therefore $Y_0$ is a compactification of $X_0$.
Okay...Is this suppose to show that $X$ has a compactification? $Y_0$ is a compactification of $X_0 = h(X)$, but why does this imply $X$ has a compactification?
We now construct a space $Y$ containing $X$ such that the pair $(X,Y)$ is homeomorphic to the pair $(X_0,Y_0)$. Let us choose a set $A$ disjoint from $X$ that is in bijective correspondence with the set $Y_0-X_0$ under some map $k : A \to Y_0 - X_0$.
Where is this set coming from? Why can it be put into bijective correspondence with $Y_0-X_0$? Are we really just taking $A = Y_0-X_0$ and taking $k$ to be the identity function? If so, why not just say that?
Define $Y = X \cup A$, and define a bijective correspondence by $H : Y \to Y_0$ by the rule $H(x) = h(x)$ for $x \in X$, and $H(a) = k(a)$ for $a \in A$.
Okay. Is $Y$ the compactification of $X$? Why is it both compact and Hausdorff? I don't see it.
In the lemma it is passed by but it is indeed possible just to do it with $A=Y_0-X_0$ and to prescribe $H:Y=X\cup A$ by $x\mapsto h(x)$ if $x\in X_0$ and $x\mapsto x$ otherwise.
After correct topologizing $Y$ (by stating that $U$ is open in $Y$ if and only if $H(U)$ is open in $Y_0$) the map $H:Y\to Y_0$ is a homeomorphism. That means that $Y$ has the same topological properties as $Y_0$. Evidently $Y_0$ is compact Hausdorff, so we are allowed to conclude that $Y$ is compact Hausdorff as well. Further (also on base of the fact that $H$ is a homeomorphism) $Y_0$ is a compactification of $X_0$ so $Y$ is a compactification of the corresponding $X\subseteq Y$.