I know that there are many posts regarding the fact that $\sin\frac{1}{x}$ is not uniformly continuous, however I am looking for help showing this directly with the $\epsilon - \delta$ definition of continuity. That is, I'm looking to show that given $\epsilon > 0$, I can find distinct $x,y$ satisfying $\lvert x - y \rvert < \delta$, but $\lvert \sin\frac{1}{x} - \sin\frac{1}{y} \rvert \geq \epsilon$, no matter which $\delta$ I choose.
I am choosing to let $\epsilon = 1$. I know that values of $x$ for which $\sin\frac{1}{x} = 1$ will be $x=\frac{1}{2\pi n + \frac{\pi}{2}}$, and values of $y$ for which $\sin\frac{1}{y} = -1$ will be $x=\frac{1}{2\pi n + \frac{3\pi}{2}}$, for $n \in \mathbb{Z}$.
However, I am struggling to determine how exactly to choose $n$. I know intuitively that as the function approaches $0$ from the right, it goes through infinitely many periods, and eventually those periods will become shorter than $\delta$, but I do not see how to translate that into a choice of $n$.
Hint
Take $\epsilon=1$ . let $\delta>0$ given and
$$x_n=\frac{1}{2n\pi+\frac{\pi}{2}}$$
$$y_n=\frac{1}{2n\pi+\frac{3\pi}{2}}.$$
as
$$\lim_{n\to +\infty}x_n=\lim_{n\to+\infty}y_n=0,$$
$$\exists N_1 \in\mathbb N : \forall n>N_1 \;\;|x_n|<\frac{\delta}{2}$$
$$\exists N_2\in\mathbb N : \forall n>N_2\;\; |y_n|<\frac{\delta}{2}$$
$\implies$
$\exists N=sup(N_1,N_2)\in \mathbb N : $
$\forall n>N \;\; |x_n-y_n|< |x_n|+|y_n|< \delta$
but
$$|f(x_n)-f(y_n)|=2>\epsilon,$$
thus $f$ is not uniformly countinuous in any interval of the form $(0,a]$.