Show that $S=\mathbb{Q} \cap [0,2]$ is not compact Question: Let $S$ be the set of rational numbers in the interval $[0, 2]$. Using the definition of compactness, show that $S$ is not compact.
Using the definition of closeness/sequential-compactness, it is easy to show that $S$ is not closed/sequential-compact, and thus it is not compact. 
But I am trying to show non-compactness of $S$ by a counterexample using purely definition of compactness, but it is not possible; since every covering example for $A=[0, 2]$ seems to be the same for $S=\mathbb{Q} \cap A$.
I highly appreciate some guidance.
EDIT Answers in here are either relevant to Topology or using non-closeness as an intermediate.         
 A: Although there is nothing wrong with proofs that do, I tried not to reference irrational numbers in the following proof.

Define a Sequence of Squares Converging to $\boldsymbol{2}$
Consider the sequence
$$
a_1=1\quad\text{and}\quad a_{n+1}=\frac{a_n}2+\frac1{a_n}\tag{1}
$$
If $a_n\in[1,2]$, then $(1)$ implies $a_{n+1}\in[1,2]$.
Each $a_n$ is in $\mathbb{Q}\cap[0,2]$ and
$$
\begin{align}
a_{n+1}^2-2
&=\frac{a_n^2}4-1+\frac1{a_n^2}\\
&=\left(\frac{a_n^2-2}{2a_n}\right)^2\tag{2}
\end{align}
$$
For $x\gt0$, $\dfrac{x^2-2}{2x}=\dfrac x2-\dfrac1x$ is monotonically increasing. Therefore, for $a_n\in[1,2]$ we have
$$
-\frac12\le\frac{a_n^2-2}{2a_n}\le\frac12\tag{3}
$$
Combining $(2)$ and $(3)$ gives
$$
\begin{align}
\left|a_{n+1}^2-2\right|
&=\left(\frac{a_n^2-2}{2a_n}\right)^2\\
&=\left|\frac{a_n^2-2}{2a_n}\right|\frac1{2a_n}\left|a_n^2-2\right|\\
&\le\frac14\left|a_n^2-2\right|\tag{4}
\end{align}
$$
which implies
$$
\lim_{n\to\infty}|a_n^2-2|=0\tag{5}
$$

Define an Infinite Cover of $\boldsymbol{\mathbb{Q}\cap[0,2]}$
Consider the open sets
$$
U_n=\left\{x\in\mathbb{Q}\cap[0,2]:|x^2-2|\gt\frac1n\right\}\tag{6}
$$
According to this answer there is no rational number so that $x^2=2$; therefore, we have
$$
\bigcup_{n=1}^\infty U_n=\mathbb{Q}\cap[0,2]\tag{7}
$$
Since $\bigcup\limits_{n=1}^mU_n=U_m$, the union of any finite collection of $U_n$ is another $U_n$. $(5)$ guarantees that no matter how big we choose $n$, there is an $a_{k_n}\not\in U_n$. Since $a_{k_n}\in\mathbb{Q}\cap[0,2]$, there is no $U_n$ that contains all of $\mathbb{Q}\cap[0,2]$.
That is, no finite subset of $\{U_n\}$ can cover all of $\mathbb{Q}\cap[0,2]$. Therefore,
$$
\mathbb{Q}\cap[0,2]\text{ is not compact.}\tag{8}
$$
A: Disclaimer: Somehow I've missed the comment of David Mitra; as my answer is essentially the same, I'm making this CW. I'm not deleting because here $U_i\cap U_j = \varnothing$, so peraps it is more clear that no finite cover exists (all the sets $U_n$ have to be taken).

Take any strictly decreasing sequence $(a_n)_{n=1,2,\ldots}$ of non-rational numbers that tends to $\sqrt{2}$ with $a_1 < 3$, for example
$$a_n = \sqrt{2}+\frac{1}{n+2015}$$
Now consider a sequence of open sets:
\begin{align}
U_0 &= (-1,\sqrt{2})\\
U_1 &= (a_1,3)\\
U_n &= (a_n,a_{n-1}) &\text{ for } n \geq 2
\end{align}
Obviously $\mathbb{Q}\cap [0,2] \subset \bigcup_{k=1}^{\infty} U_k$, but no finite cover exists.
I hope this helps $\ddot\smile$ 
A: A compact set has to be bounded and closed. We can easily see that closedness is violated here. For example,
take the famous Leibniz formula $$\frac{\pi}{4}= 1-\frac13+\frac15-\frac17+\frac19-\cdots$$ The sequence of partial sums are all rational numbers in between 0 and 2, however their limit point is not rational, so the set $S$ is not closed.
