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.         

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 $\ddot\smile$ 
