Set of rational numbers bounded between two irrationals is a closed set? Consider the metric space $\mathbb{R}$ equipped with the standard distance metric. Let $S$ be a set of rational numbers in the open interval $(a,b)$ where $a$ and $b$ are irrational. Prove that $S$ is closed in the set of rational numbers $\mathbb{Q}$.
Attempt:  
$S^c = \{x \in \mathbb{Q} : x \leq a \} \cup \{x \in \mathbb{Q} : x \geq b \}$  
Let $x < a$. Choose $\epsilon = a-x$. We can find a $N$ such that $1/N < \epsilon$.
This implies that $x + 1/N$ is rational and less than $a$. A similar argument can be made for $x > b$.
So we see, that $\forall x \in S^c, \exists \epsilon > 0$ such that $B_{\epsilon}(x) \subset S^c$. So $S^c$ is open and so $S$ is closed. 
Does this make sense? 
 A: One thing that I'd adjust is this: you never used the fact that $a,b$ are irrational. In particular, that lets you conclude that $a,b\notin S^c,$ so that $$S^c=\{x\in\Bbb Q:x<a\}\cup\{x\in\Bbb Q:x>b\}.$$ (Do you see why this is important?)
Another thing I'd adjust is the part with $N$--it doesn't really help you show that $B_\epsilon(x)\subseteq S^c.$ Rather, take $\epsilon=a-x$ as you did, and take any $y\in B_\epsilon(x),$ meaning that $y\in\Bbb Q$ and $|y-x|<\epsilon.$ In particular, since $y-x\le|y-x|,$ it then follows that $y<a.$ (Do you see how?) Consequently, $y\in S^c,$ and since $y\in B_\epsilon(x)$ was arbitrary, then $B_\epsilon(x)\subseteq S^c,$ as desired.
A: Here is another proof of this.

Theorem. Let $(M,d)$ be a metric subspace of the metric space $(N,d)$. $U \subseteq M$ is open (closed) in $M$ if and only if there is an open (closed) set $V\subseteq N$ such that $U=M \cap V$.

Using this theorem for your example, take $M=\mathbb{Q}$, $N=\mathbb{R}$, $d=|\cdot|$. Furthermore, $V=[a,b]$ is closed in $\mathbb{R}$. Also, $U=\{y\,|\,y\in\mathbb{Q}\,\text{and}\,y\in(a,b)\}$. We see that $U=\mathbb{Q}\cap V$ because $a$ and $b$ are irrational numbers. So $V$ is closed in $\mathbb{Q}$. If you take $V=(a,b)$ then similarly it is concluded that $U$ is open in $\mathbb{Q}$. This gives us that $U$ is clopen in $\mathbb{Q}$. Interestingly, $U$ is neither open nor closed in $\mathbb{R}$. Can you imagine why?
