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
You need to show that $\gamma(U)$ contains a neighborhood of $\infty$. Such a neighborhood can be of the form $\infty\cup \mathbb{C}-\bar B(0,r)$, where $\bar B(0,r)$ is the closed ball of radius $r$.
Since the topology of $\mathbb{P}^1$ is the quotient topology, $U$ contains a ball $p(B((0,1),s)$ where $p:\mathbb{C}^2-\{0\}\rightarrow \mathbb{P}^1$ is the quotient map. Let $z\in \mathbb{C}-\bar B(0,s)$, ${1\over{|z|}}<s$ implies that $({1\over z},1)\in B((0,1);s)$, this implies that $\mathbb{C}-\bar B(0,s)\subset\gamma(p(B(0,s))\subset\gamma(U)$.