How prove that $\mathcal{T}^\infty=\mathcal{T}\cup\{U:X\setminus U$ is compact in $X\}$ on $X^\infty=X\cup\{\infty\}$ is a topology
Solution 1:
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.
Solution 2:
Let us start by using closed and compact sets. By complements, I mean complements in $X$.
$\emptyset\in\tau\subset\tau^\infty$, and also closed and compact, so $X\cup\{\infty\}\in\tau^\infty$. Suppose $\mathcal{J}=\mathcal{J}_1\cup\mathcal{J}_2$, where $U_j\in\tau$ for $j\in\mathcal{J}_1$ and $U_j\in\tau^\infty\setminus\tau$ for $j\in\mathcal{J}_2$. Clearly, $\bigcup_{j\in\mathcal{J}_1}U_j\in\tau\subseteq\tau^\infty$, as $\tau$ is a topology. Moreover, by de Morgan, $\left(\bigcup_{j\in\mathcal{J}_2}U_j\right)^c=\bigcap_{j\in\mathcal{J}_2}U^c_j$ is closed and compact, so $\bigcup_{j\in\mathcal{J}_2}U_j\in\tau^\infty$, and so $\bigcup_{j\in\mathcal{J}}U_j\in\tau^\infty$. Similarly, if $|\mathcal{J}|<\infty$, we see that $\bigcap_{j\in\mathcal{J}_1}U_j\in\tau\subseteq\tau^\infty$, and $\bigcup_{j\in\mathcal{J}_2}U_j^c$ is compact and closed, so $\bigcap_{j\in\mathcal{J}_2}U_j\in\tau^\infty$. One can then check that the intersection of these intersections is either empty or in $\tau$, and hence in $\tau^\infty$.
Finally, if $X$ is Hausdorff, then compact sets are closed and your book's stuff coincides with that of Kelly's.