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?
 A: Hint:
What can you say about the sequence (and its subsequences) 
$$\left\{\left(1-\frac1n\right)^n\right\}_{n\in\Bbb N}\subset A\;\;?$$
A: Suppose there was a finite subcover. Then $A \subset (-\infty, \sqrt2 -\frac1n) \cap \mathbb{Q}$ for some $n$. However, $[\sqrt2 -\frac1n, \sqrt{2} ] \cap \mathbb{Q} \neq \emptyset$, which contradictions the definition of $A$.
A: Not sure, whether you already know that images of compact spaces under continuous maps are compact, but in case you do, this criterion may be sometimes easier to use than coming up with open covers. In particular, any continuous function on a compact attains its extremal values. Let $X := \mathbb Q\cap[0,\sqrt2]$ and define $\iota:X\to \mathbb R$ be the embedding map. Clearly, it is continuous and clearly $\sup_{x\in X}\iota(x) = \sqrt 2$, however for no $x\in X$ it holds that $\iota(x) = \sqrt2$, so $X$ is not compact.
Perhaps, in the current situation it is not much different from other solutions provided, however I think this trick is just worth keeping in mind for the future problems in topology.
A: It is not sequential compact hence not compact
$x_n={[\sqrt{2}n]\over n}\in [0,\sqrt{2}]\cap\mathbb{Q},x_n\to\sqrt{2}\notin [0,\sqrt{2}]\cap\mathbb{Q}$
A: Let $O_{n_1},\ldots,O_{n_k}$ be a finite subcover, with $n_1<\cdots<n_k$. Then
$$\left(\sqrt2-\frac{1}{n_k},\sqrt2-\frac{1}{n_k+1}\right)\cap \mathbb Q$$
is nonempty and contained in $[0,\sqrt2]\cap \mathbb Q$ but disjoint from $O_{n_1}\cup\cdots\cup O_{n_k}$, so $[0,\sqrt2]\cap \mathbb Q\nsubseteq O_{n_1}\cup\cdots\cup O_{n_k}$.
A: 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.
