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).
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,
(ii) $\langle Y,\mathcal T'\rangle$ is a compact topological space,
(iii) $\mathcal T$ is the subspace topology on $X$ induced by $\mathcal T'$, and
(iv) $X$ is not a closed subspace of $Y$ under $\mathcal T'$ (so $Y$ is the closure of $X$ under $\mathcal T'$).
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.
Best Answer
I think you are reading the wrong part of your book. That definition (top of p. 185 in my edition) is immediately preceded by a Theorem 29.1, which is a long explanation of how one can take a noncompact locally-compact Hausdorff space $X$ and embed it into a compact space $\def\Y{X^\ast}\Y$ that has exactly one point more than $X$. This larger space $\Y$ is the one-point compactification of a space $X$, sometimes called the Alexandroff compactification of $X$.
The one-point compactification $\Y$ consists of $X\cup \{\infty\}$, where $\infty$ is some new point that is not an element of $X$. The topology is as follows:
Theorem 29.1 shows that if $X$ is a Hausdorff space that is locally compact but not compact, then $\Y$, with the topology described above, is a compact Hausdorff space of which $X$ is a subspace. Theorem 29.1 also shows that any one-point compactification of $X$ must be homeomorphic to the $\Y$ described above, so that the one-point compactification of $X$ is essentially unique. This is the construction Munkres wants you to consider.
This construction is the topological formalization of the idea of taking an infinite space $X$ and "adding a point at infinity". We do this, for example, with the complex numbers, to obtain the Riemann sphere. (Ignore this example if you don't know about the Riemann sphere.) A simpler example is that the one-point compactification of $\Bbb R$ is (homeomorphic to) $S^1$, the circle: the two ends at infinity are brought together and joined at the new point $\infty$.
I hope the question makes more sense in this light.
A word of advice: In some subjects, and on standardized tests, you can read the question first, then go back and skim the material looking for something pertinent, and then answer the question without reading all the material. In advanced mathematics, this strategy will not work. You have to adopt a different strategy. First read over the entire chapter, very slowly, taking time to understand and digest each sentence before you move on to the next one. This may take several days, or more. Then do the exercises.