It's not within the sine function that dummy variables apply, it's the fact that the variables are inside an integral. Would you agree that, say: $\sum_{k=1}^n k$ is no different to $\sum_{j=1}^n j$? They both represent $1 + 2 + \cdots + n$, just with different letters. When the letters don't matter, we call them 'dummy' variables.

In the same vein, the variable you integrate with respect to in an integral doesn't matter. Is $\int_0^1 x \, \mathrm{d}x$ different to $\int_0^1 y \, \mathrm{d}y$? Nah, they both integrate to some number in the end, it doesn't matter what letter I use within it. So it's quite common to have an integral $\int_a^b f(x) \, \mathrm{d}x$, use a substitution (say $u = 2x$ for example) to get the integral $\int_{2a}^{2b} \frac{f(u/2)}{2} \, \mathrm{d}u$. But we just said that the variable we use within this integral doesn't matter, I could use any letter and it wouldn't make any difference. So I could perfectly well write it as $\int_{2a}^{2b} \frac{f(x/2)}{2} \, \mathrm{d}x$.

This situation is precisely what is going on here. You have the integral $\int_0^1 xf(\sin x) \, \mathrm{d}x$, then use the substitution $y = \pi - x$ to end up with $\int_0^1 (\pi - y) f(\sin y) \, \mathrm{d}y$. That is, you have just shown that $$\int_0^{\pi} x f(\sin x) \, \mathrm{d}x = \int_0^1 (\pi - y)\sin y \, \mathrm{d}y$$

But again, the variable within the integral is of no import, so we may as well change the $y$ to an $x$, it's a dummy variable, it doesn't matter what we call it. So we get $$\int_0^{\pi} xf(\sin x) \, \mathrm{d}x =\int_0^1 (\pi - x) f(\sin x) \, \mathrm{d}x$$

Now we can expand the second integral, and I'm sure you can follow through the rest of the proof. The important thing to note here is that variables within an integral are dummy variables and can always be swapped out for another letter if you wish. To re-iterate:

$$\int_a^b g(t) \, \mathrm{d}t = \int_a^b g(m) \, \mathrm{d}m = \int_a^b g(z) \, \mathrm{d}z = \int_a^b g(\sigma) \, \mathrm{d}\sigma = \int_a^b g(\zeta) \, \mathrm{d}\zeta = \int_a^b g(\mathrm{elephant}) \, \mathrm{d}(\mathrm{elephant})$$

This seems to be a major sticking point for students, I can't really tell why. Hope this helped. Let me know if any bit of this is unclear.