Showing $a_n=\sin(n)$ does not converge 
Show that $a_n=\sin(n)$ does not converge


My idea:
Take two subsequences: $a_{n_k}=\sin(\frac {\pi k} 2)$ , $a_{n_l}=\sin(\frac {2\pi l} 3)$
So: $\forall n$ : $\lim_{n\to\infty} a_{n_k}=1$, $\lim_{n\to\infty} a_{n_l}=-1$
So two inifinite subsequence converge to different limits thus the sequence $a_n$ doesn't converge. 
Is that correct ?
Edit: 
The contrapositive of the defintion of a limit of a sequence: 
$\exists\epsilon>0 : \forall n\in N : \exists n>N \Rightarrow |x_n-L|>\epsilon$
Take $\epsilon =1$ and we know that $\sin(n)$ is bounded so lets take it's supermum: 1
$|1-1|>\epsilon=1\Rightarrow 0>1 \Rightarrow$ Contradiction.
 A: This needs a deeper result.  Let the fractional part of $x$ be denoted by $\{x\}$. By Kronecker's Theorem, the set $\left\{ \{\frac{n}{2\pi}\}:n\in\Bbb{N}\right\}$ is a dense subset of $[0,1]$, (because $\pi$ is irrational.) So, there exist two sequences of integers $(n_k)_k$ and $(m_k)_k$ such that
$$
\lim_{k\to\infty}\left\{\frac{n_k}{2\pi}\right\}=0,\quad\hbox{and}\quad\lim_{k\to\infty}\left\{\frac{m_k}{2\pi}\right\}=\frac{1}{4}
$$
Equivalently
$$
\lim_{k\to\infty}\left(n_k-2\pi\left\lfloor\frac{n_k}{2\pi}\right\rfloor\right)=0,\quad\hbox{and}\quad\lim_{k\to\infty}
\left(m_k-2\pi\left\lfloor\frac{m_k}{2\pi}\right\rfloor\right)=\frac{\pi}{2}
$$
That is $\lim\limits_{k\to\infty}\sin(n_k)=0$ and  $\lim\limits_{k\to\infty}\sin(m_k)=1$. Thus the sequence $(\sin(n))_n$ does not converge.
$\bf{Remark.}$ A variation on this proof shows that the set $\{\sin n:n\in\Bbb{N}\}$ is dense in the interval $[-1,1]$.
A: Your argument is not correct since you compute values $\sin{k\pi\over 2}$ therein, which are not values of the original sequence.
Here is a proof that does not make use of density arguments:
Assume $\lim_{n\to\infty}\sin n=\sigma\in{\mathbb R}$. Then
$$2\cos n\>\sin 1=\sin(n+1)-\sin(n-1)\to 0\qquad(n\to\infty)\ ,$$
which implies $\lim_{n\to\infty}\cos n=0$, whence $\sigma\in\{-1,1\}$. Letting $n\to\infty$ in 
$$\sin(n+1)=\sin n\>\cos 1+\cos n\>\sin 1$$
would then imply $\cos 1=1$, which is clearly wrong.
