Prove $\bar{\mathbb{Q}}$ in $\mathbb{R}$ is $\mathbb{R}$ Take $\mathbb{R}$ in the standard (order) topology.  Show that the closure of $\mathbb{Q}$ is $\mathbb{R}$.
I'm new to topology, self-studying using the Munkres book.  In fact I'm new to proofs.  In the book he has just defined a closed set, and the notion of a closure of a set.  Here I attempt to do this proof.  I find it difficult not to appeal to things that seem 'obvious' in the course of a proof (I flag a couple of these points below).  But sometimes 'obvious' things are hard to prove indeed!  How can I improve my proof?  How would you prove it using only relatively elementary facts? 

Let us consider some irrational $x \in \mathbb{Q}'$.  For convenience, take $x>0$.  I intend to show that for any such $x$, there's no neighborhood of $x$ that does not intersect $\mathbb{Q}$, so all such $x$ have to be in the closure in question, $\bar{\mathbb{Q}}$
A minimal open set $U$ that contains $x$ is $(r_1, r_2)$ with $r_1<x<r_2$ and $r_1,r_2\in \mathbb{R}$.  But I shall show that no matter what we take for $r_1$ and $r_2$,
$$r_1<r_2 \iff \exists q \in \mathbb{Q}: r_1<q<r_2$$
To begin, note $r_1 < r_2 \iff 0<r_2-r_1$.
Let us define $\Delta r \equiv r_2 - r_1$
Consider $(\Delta r)^{-1}$ -- it could be large if $\Delta r$ is small, but certainly$^1$ $\exists z \in \mathbb{Z}_+:z>(\Delta r)^{-1}$ and this implies that ${1 \over z}<\Delta r$.  By definition, ${1 \over z} \in \mathbb{Q}$.
Next consider $m {1 \over z}$ for $m \in \mathbb{Z}$.  If we take $m=0$, then $m {1\over z}<r_1$ 
(Otherwise we get $r_1 < 0 < r_2,$ and $0 \in \mathbb{Q}$).
Also, if $m$ is large enough, then $m {1 \over z} > r_1$.  Let $M=\{0\} \cup \{m \in \mathbb{Z}_+:m {1 \over z}<r_1\}$.  Clearly $M$ is bounded by $zr_1$.  Take $n$ to be the largest element$^2$ of $M$.  Since $n \in M$, $n{1 \over z} < r_1$.  Also, $(n+1){1 \over z}>r_1$, since if not, $(n+1) \in M, (n+1)>n$ which contradicts that n is the largest element in $M$.
Now 
$$n{1 \over z} < r_1 \wedge {1 \over z} < r_2 - r_1 \implies n{1\over z} + {1 \over z} < r_1 + (r_2 - r_1) \\
\implies (n+1){1 \over z} < r_2$$
Therefore, we have that 
$$ r_1 < {n+1 \over z} < r_2$$
But of course ${n + 1 \over z} \in \mathbb{Q}$ so this neighborhood of $x$ intersects $\mathbb{Q}$ no matter what interval $(r_1, r_2)$ containing $x$ we take, assuming $x$ is positive and irrational.  This means that all neighborhoods of positive irrational numbers intersect $\mathbb{Q}$.  I believe the argument for negative irrational numbers would go the same way without much different.  This would lead to all neighborhoods of positive and negative irrational numbers intersecting $\mathbb{Q}$, thus putting them in $\bar{\mathbb{Q}}$.  But the set of all positive or negative irrationals plus all rationals is itself $\mathbb{R}$ so we have $\bar{\mathbb{Q}} = \mathbb{R}$.
How did I do?
$^1$ I didn't show that $\forall r \in \mathbb{R}: \exists q \in \mathbb{Z}_+:q>r$.  I find it difficult to prove something that seems so obvious.  But I would like to see it proved once.  How can we get that?
$^2$ I didn't show that $M$ has a largest element, how might I do that?
 A: Let $x<y$. Then choose $n$ such that $\frac{1}{n} < y-x$. Now note that we must have $\{\frac{m}{n}\}_m \cap (x,y) \neq \emptyset$. Hence any non-empty open set contains a rational.
Now consider $\overline{Q}$, which is closed by definition. Hence $(\overline{Q})^C$ is open. However, it cannot contain a rational, so it must be empty.
Addendum: To show that $\{\frac{m}{n}\}_m \cap (x,y) \neq \emptyset$, note that $\mathbb{R} = \cup_m [\frac{m}{n}, \frac{m+1}{n})$, and that the union is disjoint. Hence for some $m$, we have $[\frac{m}{n}, \frac{m+1}{n}) \cap (x,y) \neq\emptyset$. If $\frac{m}{n} \in (x,y)$ we are finished, so suppose $\frac{m}{n} \not\in (x,y)$. Then we must have $\frac{m}{n} \leq x <\frac{m+1}{n}$ (otherwise there is no overlap). Since $\frac{1}{n} < y-x$, we have $\frac{m}{n} + \frac{1}{n} = \frac{m+1}{n} < y$ and so $\frac{m+1}{n} \in (x,y)$.
A: It would be easier to do this. Pick $x,y$ reals. Assume $x>y$, so $x-y>0$. Archimedes says there is $n$ such that $n(x-y)=nx-ny>1$. Can you find an integer between $nx$ and $ny$ ${}^{1}$? Observe that $nx-ny>1$, which is important! If so, can you find a rational between $x$ and $y$? Why does this tell you that $\overline{ \Bbb Q}=\Bbb R$?

$1$. Spoiler Consider $m=\lfloor ny\rfloor+1$. Then by definition 

$$ny<m\leq ny+1<ny+nx-ny=nx$$ 


DEF Let $x$ be a real number. We define $\langle x\rangle$ to be the greatest element of $$S=\{m\in\Bbb Z:m\leqslant x\}$$
This set is nonempty by the Archimedean property. In particular, we can reduce it to a finite set by choosing $m'$ such that $m'<x-1$. Then $m'\notin S$, and we may look at $$S=\{m\in\Bbb Z:m'< m\leqslant x\}$$
This is a non-empty finite set of integers, thus it admits a maximum (unique) element.
