Edit: Now that your post has been fixed, let me respond appropriately. First, let me clarify a few points.
In fact, every non-compact set admits a one-point compactification, in the following sense:
Proposition: Suppose $X$ a set, $\mathcal T$ a topology on $X$ such that $\langle X,\mathcal T\rangle$ is not a compact space. Then there is a set $Y\supseteq X$ and a topology $\mathcal T'$ on $Y$ such that:
(i) $Y\smallsetminus X$ consists of a single point,
<p>(ii) $\langle Y,\mathcal T'\rangle$ is a compact topological space,</p>
<p>(iii) $\mathcal T$ is the subspace topology on $X$ induced by $\mathcal T'$, and</p>
<p>(iv) $X$ is not a closed subspace of $Y$ under $\mathcal T'$ (so $Y$ is the closure of $X$ under $\mathcal T'$).</p>
In particular, taking some object not in $X$--call it $\infty$--letting $Y=X\cup\{\infty\},$ and letting
$\mathcal T':=\mathcal T\cup\{Y\smallsetminus K:K\subseteq X\text{ is compact }\textit{and closed}\text{ in }X\text{ under }\mathcal T\},$
we have that $Y,\mathcal T'$ thus constructed satisfy the given conditions. We call such a space $\langle Y,\mathcal T'\rangle$ a one-point compactification of $\langle X,\mathcal T\rangle.$ If $\langle X,\mathcal T\rangle$ has the property that compact sets are closed (for example, if it is Hausdorff), we don't need the "and closed" requirement in the definition of $\mathcal T'$.
Moreover, if $\langle Y,\mathcal T'\rangle$ and $\langle Z,\mathcal T''\rangle$ are both one-point compactifications of a locally compact Hausdorff space $\langle X,\mathcal T\rangle$, then they are homeomorphic (so in such a circumstance, it makes sense to talk about the one-point compactification of a topological space, rather than a one-point compactification). [To see why the original space must be locally compact Hausdorff for uniqueness, see Brian's answer here.]
We also have the following:
Corollary: Suppose $X$ a set and $\mathcal T$ a topology on $X$ such that $X$ is not compact under $\mathcal T$. $\langle X,\mathcal T\rangle$ is locally compact Hausdorff if and only if it has a (unique) one-point compactification that is Hausdorff.
The above are good exercises to prove.
Now, the rest depends on what you mean by $X\cup\{(1,1)\}$.
If you intend that $X\cup\{(1,1)\}$ be simply considered as a subspace of $\Bbb R^2$ (which it seems that you do), then we can construct the one-point compactification of $X\cup\{(1,1)\}$ by the Proposition. However, it will not be a Hausdorff space. Note that for any neighborhood $U$ of $(1,1)$ in $X\cup\{(1,1)\}$, and for any $K$ such that $U\subseteq K\subseteq X\cup\{(1,1)\},$ we have that $K$ has as part of its boundary (in $\Bbb R^2$) the open segment from $(1,1)$ to $(\alpha,1)$ for some $0<\alpha<1$, and this open segment is disjoint from $K$, so in particular, $K$ is not compact. (Why not?) Thus, $X\cup\{(1,1)\}$ is not locally compact (though it is Hausdorff), so by the Corollary, its one-point compactification is not Hausdorff (though the compactification does exist by the Proposition, and is $T_1$). Now, the one-point compactification of $X$ is homeomorphic to the closed triangular figure with vertices $(0,0),(1,0),(1,1)$. The one-point compactification of $X\cup\{(1,1)\}$ is the quotient space $$[0,1]^2/\bigl\{(x,1)\in[0,1]^2:x\ne 1\bigr\}.$$ I can't see any "nice" spaces homeomorphic to that. In particular (as kahen points out in his comment below), it isn't a pseudometrizable space. It does have some nice properties, though, such as uniqueness of sequence limits (as I lay out in the comments below), which not all $T_1$ spaces satisfy.
If, on the other hand, you intend that $X\cup\{(1,1)\}$ be topologized as a disjoint union--which in particular means that $(1,1)$ is an isolated point--then the one-point compactification of $X\cup\{(1,1)\}$ is simply the one-point compactification of $X$, together with another point that is isolated.
\cup {(1,1)}obviously intending\cup \{(1,1)\}. – kahen May 07 '13 at 01:39