Proving that $\{\cos n\mid n\in\mathbb N\}$ is dense in $[-1,1]$ I am trying to prove that $\cos n$, $n\in \mathbb{N}$,  is dense in $[-1,1]$ and would like to know if my answer is rigorous and correct enough. 

Given any $x,y \in [-1,1]$ with $y>x$, there are $a,b \in [0,\pi)$ with $a>b$ such that $\cos a = x$ and $\cos b = y$, since the cosine function is decreasing on $[0,\pi)$.
  Using the fact that $\{m+n\pi\mid m,n\in\mathbb{Z}\}$ is dense on $\mathbb{R}$, there exists $m+2n\pi$ $\in(b,a)$. Hence there is a $\cos(m+2n\pi)=\cos (m)\in (y,x) $ with $m\in \mathbb{N}$.

Is it correct?
 A: Arguments don't just exist in vacuums waiting to be determined "correct" or otherwise. In this case, comment by HCP16 is appropriate: If you are willing to assume that $\{m + n\pi\}_{m,n \in \mathbb{Z}}$ is dense in $\mathbb{R}$, then a proof like this ought to work easily. 
Taking what you've written at face value, one question that comes to my mind is: How do you go from $m + n \pi$ being dense to $m + 2n\pi \in (b,a)$?
A: As others have said, how good your argument is depends greatly on what you are assuming has already been shown, but there are a couple of genuine minor gaps. If you are willing to assume that the cosine function is continuous and is decreasing on $[0,\pi]$, you can appeal to the intermediate value theorem to justify the existence of your $a$ and $b$, but you do need to use the closed interval $[0,\pi]$, not $[0,\pi)$: otherwise you can’t handle the possibility that $x=-1$.
If you are willing to assume that $\{m+n\alpha:m,n\in\Bbb Z\}$ is dense in $\Bbb R$ when $\alpha$ is irrational, then certainly $\{m+n\pi:m,n\in\Bbb Z\}$ is dense in $\Bbb R$, but that does not obviously justify your conclusion that there are $m,n\in\Bbb Z$ such that $m+2n\pi\in(b,a)$. However, $2\pi$ is also irrational, so $\{m+2n\pi:m,n\in\Bbb Z\}$ is also dense in $\Bbb R$, and we can indeed conclude that there are $m,n\in\Bbb R$ such that $m+2n\pi\in(b,a)$. (And I now see that you corrected that oversight in the comments, though not yet in the question itself.)
However, the fact that $\{m+n\alpha:m,n\in\Bbb Z\}$ is dense in $\Bbb R$ when $\alpha$ is irrational is really the heart of the argument and its most difficult part1: the rest is comparatively trivial, assuming that the intermediate value theorem and basic facts about the cosine function are already available (and I do think that a reasonable assumption). I would therefore expect a proof of the desired result to include (or, in a textbook, directly refer to) a proof of this fact. 
That can actually be quite short: in this answer I used the pigeonhole principle to give a short, easy proof that $\{n\alpha\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$ if $\alpha$ is irrational, where $x\bmod 1=x-\lfloor x\rfloor$ is the fractional part of $x$, and from this it follows immediately that $\{m+n\alpha:m,n\in\Bbb Z\}$ is dense in $\Bbb R$. In this answer I point out that the same argument works if we replace $\Bbb Z$ by $\Bbb Z^+$ and show how to extend the argument to show that in fact $\{m+n\alpha:m,n\in\Bbb Z^+\}$ is dense in $\Bbb R$.
1 The proof that $\pi$ is irrational is actually harder, but in most contexts that fact qualifies as known even if no proof has been exhibited. 
