You get the uniqueness result if the space is Hausdorff.
Let $\langle X,\tau\rangle$ be a compact space. Suppose that $p\in X$ is in the closure of $Y=X\setminus\{p\}$, and let $\tau_Y$ be the associated subspace topology on $Y$; $\langle X,\tau\rangle$ is then a compactification of $\langle Y,\tau_Y\rangle$.
Suppose that $p\in U\in\tau$, and let $V=U\cap Y$. Then $\varnothing\ne V\in\tau_Y$, so $Y\setminus V$ is closed in $Y$. Moreover, $Y\setminus V=X\setminus U$ is also closed in $X$, which is compact, so $Y\setminus V$ is compact. That is, every open nbhd of $p$ in $X$ is the complement of a compact, closed subset of $Y$. Thus, if $\tau'$ is the topology on $X$ that makes it a copy of the Alexandroff compactification of $Y$, then $\tau\subseteq\tau'$.
Now let $K\subseteq Y$ be compact and closed in $Y$, and let $V=Y\setminus K\in\tau_Y$. If $X\setminus K=V\cup\{p\}\notin\tau$, then $p\in\operatorname{cl}_XK$. If $X$ is Hausdorff, this is impossible: in that case $K$ is a compact subset of the Hausdorff space $X$ and is therefore closed in $X$. Thus, if $X$ is Hausdorff we must have $\tau=\tau'$, and $X$ is (homeomorphic to) the Alexandroff compactification of $Y$.
If $X$ is not Hausdorff, however, we can have $\tau\subsetneqq\tau'$. A simple example is the sequence with two limits. Let $D$ be a countably infinite set, let $p$ and $q$ be distinct points not in $D$, and let $X=D\cup\{p,q\}$. Points of $D$ are isolated. Basic open nbhds of $p$ are the sets of the form $\{p\}\cup(D\setminus F)$ for finite $F\subseteq D$, and basic open nbhds of $q$ are the sets of the form $\{q\}\cup(D\setminus F)$ for finite $F\subseteq D$. Let $Y=D\cup\{q\}$. Then $Y$ is dense in $X$, and $X$ is compact, and $Y$ itself is a closed, compact subset of $Y$ whose complement is not open in $X$.
Improved example (1 June 2015): Let $D$ and $E$ be disjoint countably infinite sets, let $p$ and $q$ be distinct points not in $D\cup E$, let $X=D\cup E\cup\{p,q\}$, and let $Y=D\cup E\cup\{q\}$. Points of $D\cup E$ are isolated. Basic open nbhds of $q$ are the sets of the form $\{q\}\cup (E\setminus F)$ for finite $F\subseteq E$, and basic open nbhds of $p$ are the sets of the form $\{p\}\cup\big((D\cup E)\setminus F\big)$ for finite $F\subseteq D\cup E$. Then $Y$ is a non-compact dense subspace of the compact space $X$, so $X$ is a (non-Hausdorff) compactification of $Y$. Let $K=\{q\}\cup E$. Then $K$ is a compact closed subset of $Y$, but $X\setminus K=\{p\}\cup D$ is not open in $X$.
(This avoids the question of whether it’s legitimate to look at the Alexandrov compactification of a compact space.)
You seem to be confusing things. First $\coprod_{j \in J} X_j$ (the coproduct or topological sum) should be the "disjointified" union $$\bigcup_{j \in J} X_j \times \{j\}$$ and the injections are $\sigma_j(x)=(x,j)$ for all $j \in J$.
Then $\coprod_{j \in J} X_j$ indeed has the final topology $\mathcal{T}_f$ wrt the injections $\sigma_j:(X,\mathcal{T}_j) \to X$ as you describe. Open sets $O$ in $\coprod_{j \in J} X_j$ can be naturally written as $$ O = \bigcup_{j \in J} O \cap \sigma_j[X] = \bigcup_{j \in J} \sigma_j[\sigma_j^{-1}[O]]$$ which we can see as a union of (homeomorphic copies of) open subsets from the $X_j$.
We could also form the union $X = \bigcup_{j \in J} X_j$ and have injections $\iota_j(x)=x$ and get a final topology $\mathcal{T}'_f$ on $X$ wrt those injections. But then we need assumptions on the $X_j$ and their mutual relations to have any nice meaningful topology on $X$. E.g. suppose that $X_1=[0,1]$ in its usual topology and $X_2 = [0,1]$ in the discrete topology. Then, when executing this construction, $X \simeq X_1$. There is no trace of $X_2$ anymore. We cannot distinguish between the $\frac{1}{2}$ that comes from $X_1$ and the one from $X_2$ etc. Open sets in $X$ are not unions of open sets from $X_1$ and $X_2$ any more. This construction is never done in practice; it doesn't seem to be useful. The standard sums (first part) are very common however, mostly as a tool in other results.
One common exception where the product view can be valid: if all $X_j$ are the same topological space, say $\forall j: (X_j, \mathcal{T}_j)=(Y, \mathcal{T}_Y)$. In that case $\coprod_{j \in J} X_j \simeq Y \times J$ where $J$ has the discrete (!) topology and $Y \times J$ has the product topology. The homeomorphism is just the identity, which is well-defined on the set level. Check that it's continuous and open between these topologies..
Best Answer
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.