To apply the residue theorem to
$$\int_{|z|=1} \frac{z}{\sin 1/z}dz $$
you need to show that $$\lim_{r \to 0}\int_{|z|=r} \frac{z}{\sin 1/z}dz=0$$ obtaining
$$\int_{|z|=1} \frac{z}{\sin 1/z}dz = \lim_{N \to \infty} \int_{|z|=1} \frac{z}{\sin 1/z}dz-\int_{|z|=2\pi/(N+1/2)} \frac{z}{\sin 1/z}dz \\ = \lim_{N \to \infty} 2i\pi\sum_{n=7}^N \frac{2\pi/n}{ \frac{-1}{(2\pi /n)^2}\cos(\frac{1}{2\pi /n})}+\frac{-2\pi/n}{ \frac{-1}{(-2\pi/n)^2}\cos(\frac{1}{-2\pi /n})}$$
Here the obtained series converges absolutely but in general you need to keep the order of summation according to the annulus where each residue comes from, that $\lim_{r \to 0}\int_{|z|=r} \frac{z}{\sin 1/z}dz=0$ implies the series converges.
Very reasonable question! I wondered about this for decades, myself! :)
The integral does not literally converge. It does converge in a "Cauchy principal value" sense, but this requires that we make a convention, or do something. It is not in any way automatic, any more than $\int_{\mathbb R} f(x)/x\;dx$ "automatically" takes the "Cauchy principal value" value.
The more bare, real fact is that that integral "through" the pole is not well-defined, since, after all, as a literal integral (as opposed to something with conventions imposed) it does not converge at all.
This explains why there's no "proof" that a contour integral "through" a pole picks up half the residue. Because the assertion is not literally true, as stated. Sure, we can say something about the related principal value integral, but that is a very different thing.
(And the possibilities of other "angles" of contour through poles likewise need principal value interpretations, otherwise are not well-defined. And, NB, there is no mandate to take the PV interpretation, so, in particular, the literal integrals do not magically/automatically take those values.)
EDIT: also, in case people might too glibly assume that there's not real issue about "regularizing" such integrals, please do consider the precise assertion of the Sokhotski-Plemelj theorem (eminently google-able). That is, it turns out that it is easy to imagine false things in terms of regularization.
Best Answer
Another approach to evaluating the integral with appealing to either Cauchy's Integral Formula or the Residue Theorem relies on Cauchy's Theorem.
If $f(z)$ analytic within a open region (simply connected) in the complex plane, then for any contour (with finite length) $C$ contained in that open region,
$$\oint_C f(z) dz=0$$
Now, in the problem at hand, $f(z)$ is not analytic in the region encircled by $C$. However, we may evaluate the integral on a "deformed" contour $C'$ that does not enclose any singularity. Then,
$$\oint_{C'} f(z)dz=0$$
If we deform $C$ by adding a "key-hole" contour that "cuts out" the singularity with a small circle $\gamma$ of radius $\epsilon$, centered at the singular point then
$$\oint_{C'} f(z)dz=\oint_C f(z)dz-\oint_{\gamma}f(z)dz=0$$
where the opposing contributions from the "key length" integrations annihilate one another.
Thus, this reduced the problem to evaluating the integral over $\gamma$. For this problem,
$$\begin{align} \oint_{\gamma}f(z) dz&=\lim_{\epsilon \to 0}\left(\int_0^{2\pi}\frac{i\epsilon e^{i\phi}d\phi}{(i+\epsilon e^{i\phi})^2+1}\right)\\\\ &=\lim_{\epsilon \to 0}\left(\int_0^{2\pi}\frac{i\epsilon e^{i\phi}d\phi}{2i\epsilon e^{i\phi}+\epsilon^2 e^{i2\phi}}\right)\\\\ &=\pi \end{align}$$