A set $Y$ is discrete in $X$ iff all points of $Y$ are isolated in $Y$. So for $y\in Y$ there is an open set $O$ of $X$ such that $O \cap Y =\{y\}$, so all singletons of $Y$ are open in $Y$.
Because all subsets of $Y$ are unions of singletons, all subsets of $Y$ are open in $Y$. It follows that all subsets of $Y$ are closed in $Y$ too. And as in the lemma situation $Y$ is not only discrete but also closed in $X$, all subsets $A$ of $Y$ are also closed in $X$. ($A = C \cap Y$ for some closed $C$ in $X$, but this is an intersection of two closed sets in $X$, so closed in $X$). That’s why we have that many sets of disjoint closed subsets in $X$, every pair of disjoint sets in $Y$ will give such pairs, in particular $A$ and $Y\setminus A$ for any $A \subseteq Y$, as is used here.
So let $$\mathcal{T}^* := \{U \subseteq X^*\mid U \cap X \in \mathcal{T} \land (\infty \in U \implies X \setminus U \mathrm{\ compact)}\}$$
instead (as in the linked question) which does say something different then you had before: An open neighbourhood of the new point $\infty$ is of the form $U \cup \{\infty\}$ such that $U \subseteq X$ is already open in $\mathcal{T}$ and $X\setminus U$ is compact (in $(X,\mathcal{T})$ of course).
This is also consistent with the Spanish: you have to read it as ($U \cap X$ open in $X$) and ( either $U \subseteq X$, so $\infty \notin U$ or ( $\infty \in U$ and) $X\setminus U$ is compact ). So for any $U$ we have $U \cap X$ is open in $X$ at least. (and this is important for being a compactification, see below). The formulation as I gave it above is more explicit. It shows there are two types of open sets: the old ones from $X$ and new ones of the form $U \cup \{\infty\}$ where $U$ is "old" open and moreover has compact complement in $X$, which is important to show the new space to be compact.
Now if we have a union of such open sets $O_i, i \in I$ there are two cases:
$\infty \notin O:=\bigcup_{i \in I} O_i$. So all $O_i \in \mathcal{T}$ and hence so is their union, and $O \in \mathcal{T}^\ast$.
$\infty \in O:=\bigcup_{i \in I} O_i$, so for some $j \in I$, $\infty \in O_i$ so $O_j= O'_j \cup \{\infty\}$ with $O'_j$ open in $\mathcal{T}$ and $X\setminus O_j$ compact in $(X,\mathcal{T})$. But then $X\setminus O \subseteq X\setminus O_j$ and this is closed in that compact set (as $O \cap X \in \mathcal{T}$) and hence compact, and so $O$ also is in $\mathcal{T}^\ast$.
The two set intersection axiom is also boring: there are three cases (none of them contains $\infty$, just one of them, or both); all are easily checked. E.g. if $\infty \in O_1, O_2$, then $O_1 \cap O_2 \cap X = (O_1 \cap X) \cap (O_2 \cap X) \in \mathcal{T}$ and $X\setminus (O_1 \cap O_2) =(X\setminus O_1) \cup (X\setminus O_2)$ is a union of two compact sets, so compact.
And as $\emptyset \in \mathcal{T}$ it is in $\mathcal{T}^\ast$ too and $X^\ast$ has $X^\ast \cap X = X \in \mathcal{T}$ and $X\setminus X^\ast=\emptyset$ is compact, so $X^\ast \in \mathcal{T}^\ast$ too.
Note that for any $U \in \mathcal{T}^\ast$, $U \cap X \in \mathcal{T}$ by definition, and $\mathcal{T} \subseteq \mathcal{T}^\ast$ so that the subspace topology of $X$ as a subspace of $X^\ast$, is just $\mathcal{T}$ again. This is not garantueed to be the case if we just demand that open neighbourhoods $U$ of $\infty$ have $X\setminus U$ compact. And this is an essential property for a compactifcation of $X$! If $X$ is Hausdorff to start with, this is a non-issue, but e.g. take $X=\Bbb Z$ with the cofinite topology on $\Bbb Z^-$ and the discrete one on $\Bbb Z^+_0$ (to make $X$ non-compact) and then your book's construction makes $X$ (as a subspace of your $A^\infty$) a discrete subspace.
Best Answer
In 3) the phrase ' there exist an open set $U$' should be changed to ' there exist a non-empty open set $U$'. [Otherwise the result is false since we can always take $U$ to be the empty set].
Suppose 3) holds and $E$ is not nowhere dense. Then the interior of $\overline {E}$ is a non-empty open set $A$. [$\overline {E}$ is the closure of $E$]. By 3) there exist a non-empty open set $U$ such that $U \subseteq A$ and $U \cap E =\emptyset$. If $x$ is any point of $U$ then $x \in A \subseteq \overline {E}$. But $U$ is a neighborhood of $x$ which does not intersect $E$. This is a contradiction.