Let $\ K=C[0,1]$, the set of continuous functions on interval $[0,1]$ with the supremum norm, and consider the set $\ A= \{ x \rightarrow \sin(\pi nx) : n \in \mathbb{Z} \}$. Prove or disprove whether A is compact or not.
I am trying to prove that it is not compact by showing that it is not equicontinuous. I noticed that you can make the "period" of sine small like so $$ 0<\pi nx < 2\pi $$ $$0<x<2/n $$ That way $\delta$ cannot be made small enough. I don't know if this reasoning is correct or how to formalize it.
Arzela-Ascoli is definitely one way to go. As you say, we should show that the sequence of functions $f_n(x)=\sin(\pi n x)$ is not equicontinuous. You're also on the right track -- noting that the periods can be made arbitrarily small, but the amplitude is still 1.
Here is a formal argument. Fix any $\delta>0$. By the Archimedian property of the reals, we can choose $N$ so large that $2/N<\delta$. Then for this $N$, $$\vert 0-1/(2N)\vert =1/(2N)<2/N<\delta$$ but then for the function $f_N(x)=\sin(\pi N x)$, we have $$\vert \sin(\pi N \cdot 0)-\sin(\pi N (1/(2N)))\vert=\vert 0-\sin(\pi/2)\vert=1\;.$$ So we have shown that there exists $\epsilon>0$ (namely $\epsilon=1$) such that for all $\delta>0$, there exists two points $x,y$ with $\vert x-y\vert <\delta$ (namely, $0$ and $1/(2N)$) so that for the function $f_N(x)=\sin(\pi N x)$, we have $$\vert \sin(\pi N x)-\sin(\pi N y)\vert\geq \epsilon\;.$$ By definition, the sequence is not equicontinuous.