Finding the closure of A=$\mathbb Q \cap [0,1]$. Finding the closure of A=$\mathbb Q \cap [0,1]$.
My attempt: Between every two rational numbers, there is a rational number. That means that for  any arbitrarily small $\epsilon \gt0$, and some $q\in \mathbb Q \cap [0,1]$  there is $x \notin \mathbb Q$, $x\in [0,1]$ so that $x\in (q-\epsilon, q+\epsilon)$. That means - and I'm not sure if that's correct and how to show it - there is a sequence $\{a_n\}_{n=1}^\infty \subseteq \mathbb Q \cap [0,1]$ so that $\lim_{n \rightarrow \infty} a_n=x$. 
Therefore, the closure of A is $[0,1]$. 
Is this correct? And how do I show my missing step (if it's valid?).
Thanks for any assistance in advance! 
 A: A point $x \in [0,1]$ is in the closure of $\mathbb{Q}\cap [0,1]$ if and only if every open set $U$ containing $x$ intersects $\mathbb{Q}\cap [0,1]$.
Take an open set $U$ containing $x$. There exists an open interval $(a,b)$ such that $x \in(a,b) \subseteq U$ (why ?). This interval $(a,b)$ surely intersects $\mathbb{Q}\cap [0,1]$ (why ?).
A: For every $x \in [0,1]$ there are rationals $q$ arbitrarily close to $x$. For each natural number $n$, choose $a_n \in \mathbb{Q} \cap [0,1]$ such that $|x-a_n| < \frac1n$. Then we have $\lim_{n \to \infty} a_n = x$.
A: I am not sure if this can be called as an answer :
Here is my idea :
You want to find smallest closed set which contains $A=\mathbb{Q}\cap [0,1]$ name it $\bar{A}$
For all good reasons, I can say that your $\bar{A}$ has to be a subset of $[0,1]$ (???)
So, you have to see if some element of $[0,1]$ is missing in the closure...
Suppose $a\in [0,1]$ is missed in the closure.. then? 
I would go further and claim that if some thing is missing that can not be a rational number.. (????)
May be you can say no irrational number is missing (:D) by using density of rational numbers...
