Show that $A= ([0,\sqrt2] \cap \Bbb Q ) \subset \Bbb Q $ is not compact. We have $\Bbb  Q$ equipped with the Euclidean Metric.
Show that $A=([0,\sqrt2] \cap \Bbb Q ) \subset \Bbb Q  $ is not compact.
How would you go about showing this?
You can make on open cover $\{O_n\}= ((-\infty, \sqrt2 -\frac1n) \cap \Bbb Q)$ for $n\ge1$
$ A \subset {O_n}$
How would you show there is no finite subcover?

If a subset $A$ of $\mathbb R$ is compact, then it is also closed. In particular, if a ssequence $\{x_n\}_{n\in}\subset A$ converges to $x$, then $x\in A$. Let
\begin{align}
x_1=&1\\
x_2=&1.4 \\
x_3=&1.41 \\
x_4=&1.414 \\
x_5=&1.4142 \\
etc.
\end{align}
I.e., 
$$
x_n=\frac{\lfloor10^n\sqrt{2}\rfloor}{10^n},
$$
Clearly $x_n\to\sqrt{2}$, as $|x_n-\sqrt{2}|<10^{-n}$, and $\sqrt{2}\not\in A$. Hence $A$ is not compact.
