Prove that the following sequence $(\sin\frac{\pi k}{3})_{k=1}^\infty$ diverges $$x_k := \sin \left(\frac{\pi k}{3} \right),\qquad x_{\infty} := \lim_{k \to \infty}{x_k}$$

I tried to prove by contradiction. Assume limit exists and is $L$. Fix $\epsilon=\frac{\sqrt{3}}{4}$. There exists $K\in\mathbb{Z}^{+}$ such that $k>K$ implies $$|\sin\frac{\pi k}{3}-L|<\epsilon$$ Notice that for $k=6l+1$ or $k=6l+2$ where $l\in\mathbb{Z}^{\geq 0}$, $\lvert \frac{\sqrt{3}}{2}-L\rvert<\frac{\sqrt{3}}{4}$. Also, for $k=6l+4$ or $k=6l+5$ where $l\in\mathbb{Z}^{\geq 0}$, $\lvert -\frac{\sqrt{3}}{2}-L\rvert<\frac{\sqrt{3}}{4}$. Now we know that $\sqrt{3}=\lvert\frac{\sqrt{3}}{2}-\big(-\frac{\sqrt{3}}{2}\big)\rvert=|\frac{\sqrt{3}}{2}+L-L-\big(-\frac{\sqrt{3}}{2}\big)|\leq|\frac{\sqrt{3}}{2}+L|+|-L-\big(-\frac{\sqrt{3}}{2}\big)|$ $=|-\frac{\sqrt{3}}{2}-L|+|\frac{\sqrt{3}}{2}-L|<\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\;\Rightarrow\;\sqrt{3}<\frac{\sqrt{3}}{2}$. Hence, we get a contradiction. Therefore, the sequence diverges.
Is this proof correct? If not, can you help me to prove using epsilon-delta definition?

Let's consider 3 sub sequences
$$k=6n \Rightarrow \sin\frac{\pi k}{3}=\sin 2\pi n=0 $$
$$k=6n+1 \Rightarrow \sin\frac{\pi k}{3}=\sin \left(2\pi n +\frac{\pi }{3} \right) = \frac{\sqrt{3} }{2}$$
$$k=9n+1 \Rightarrow \sin\frac{\pi k}{3}=\sin \left(2\pi n +\pi+\frac{\pi }{3} \right) = -\frac{\sqrt{3} }{2}$$
So we have 3 limit points.
