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\noindent 
\textbf{2.12.}
Let $K$ be the following subset of $\R$: 
\[
K=\{0\} \cup \left\{\frac{1}{n}: n = 1, 2, 3, \ldots \right\}.
\]
Prove that $K$ is compact directly from the definition.

Let $\{G_{\alpha} \}$ be an open cover of $K$.  I have to show that
there is a finite subcover.  Since this is an open cover, $K \subseteq
\bigcup_{\alpha} G_{\alpha}$, so in particular, the point $0$ is in
one of the sets in the cover, say $G_{\alpha_{0}}$.  Since
$G_{\alpha_{0}}$ is open, every point in it is an interior point, and
in particular, $0$ is an interior point.  So $G_{\alpha_{0}}$ contains
a neighborhood of the form $N_{\epsilon}(0)$ for some $\epsilon > 0$.
Since $\epsilon >0$, there is an $n$ so that $\epsilon > 1/n$ (by the
archimedean property of $\R$, Theorem 1.20(a), with $x=\epsilon$ and
$y=1$).  Therefore the neighborhood contains all points $x$ on the real
line with $|x|\leq 1/n$, so the points
\[
\frac{1}{n}, \ \frac{1}{n+1}, \ \frac{1}{n+2}, \ \ldots
\]
are all contained in $N_{\epsilon}(0) \subseteq G_{\alpha_{0}}$.

Now I'm ready to exhibit a finite subcover of $\{G_{\alpha}\}$.  For
each number $i=1$, $2$, \dots, $n-1$, find an open set $G_{\alpha_{i}}$
from the cover which contains $1/i$.  Then $K$ is contained in 
\[
G_{\alpha_{0}} \cup G_{\alpha_{1}} \cup G_{\alpha_{2}} \cup \cdots
\cup G_{\alpha_{n-1}}.
\]
Why?  For $i=1, \ldots, n-1$, the point $1/i$ is in $G_{\alpha_{i}}$.
The rest of the points in $K$ are in $G_{\alpha_{0}}$.  Therefore 
\[
\{G_{\alpha_{0}}, G_{\alpha_{1}}, \ldots, G_{\alpha_{n-1}} \}
\]
is a finite subcover.  I have showed that every open cover of $K$
contains a finite subcover, and therefore $K$ is compact.

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