What we actually care about is relationships between topological spaces, and "$A$ is the one-point compactification of $B$" happens to be a particularly nice relationship about which it is possible to say a lot. For example, the sphere $S^n$ is the one-point compactification of $\mathbb{R}^n$, and this observation makes it possible to prove things about $S^n$ by passing to $\mathbb{R}^n$ or vice versa.
Example. The sphere $S^n$ is simply connected. One way to prove this is to show that a path in $S^n$ can be deformed so that it misses one point (this is the hard step). From here, removing the missed point gives a path in $\mathbb{R}^n$, which can be deformed into a constant path using linear functions.
In addition, a general principle in mathematics is that things which are unique are probably important. The one-point compactification is a construction of this type: it is the unique minimal compactification (of a locally compact Hausdorff space).
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.)
Best Answer
We are given a set $X$ with some topology. We're working in a universe of sets, so there are always sets $A$ that are not an element of $X$ (there is no universal set, by Russell's paradox, or $\mathscr{P}(X)$ is a strictly larger set than $X$, etc. There are several argument depending on your axioms, e.g. if Regularity holds, then $A=X$ itself will work, as $X \notin X$ in that case). However we get it, we have some point (or set, as everything is a set) $\infty \notin X$.
We then proceed to define a topology on the larger set $X \cup \{\infty\}$ etc.
It's somewhat similar to working inside $\Bbb R$ and declaring the existence of some abstract number $i$ that satisfies $i^2=-1$ (and so cannot be a member of $\Bbb R$) and constructing the complex numbers from that. In that case we can also find another "model" for it, by using the product $\Bbb R^2$ and seeing $i$ as $(0,1)$ etc. In topology we can also often find a model for the Alexandroff compactification of $X$ inside some other space we already know (like $\Bbb S^2 \subseteq \Bbb R^3$ for $\Bbb R^2$ and $\Bbb S^1$ for $\Bbb R$, or a "figure 8" in the plane for $(0,1) \cup (1,2)$ etc. We can then show that this concrete space is homeomorphic to the abstract construction $X \cup \{\infty\}$ we start with. But we don't need that model to define the Alexandroff compactification, it's a helpful intuition, sometimes.