Let $u=\pi x$. The integral turns into
\[\frac{1}{\pi}\int_0^{\pi}\log(\sin u)\,du.\]
Consider the function
\[f(z)=\log(1-e^{2iz})=\log(-2ie^{iz}\sin z)=\log(1-e^{-2y}(\cos(2x)+i\sin(2x))).\]
This function is real and negative if $y<0$ and $x=n\pi$. If we delete these half lines, we can assume that $\log$ is single-valued and analytic.
We now integrate over the rectangle with corners $z=0$, $z=\pi$, $z=\pi+iy$ and $z=iy$ (and then we let $y\to\infty$ as seen in figure). At the points $0$ and $\pi$, we create circular arcs of radius $\epsilon$ to avoid these points. By periodicity of the function, the integral along the vertical lines is zero. Also, note that if $L$ is the line connecting $iy$ with $\pi+iy$, then
\[\left|\int_L f(z)\,dz\right|\leq 2\pi|e^{iz}|=2\pi e^{-y}\rightarrow 0\]
as $y\rightarrow\infty$. Now, notice that the imaginary part of the logarithm is bounded. Thus, we only need to worry about the real part. Observe that
\[\left|\frac{1-e^{2iz}}{z}\right|\rightarrow 2\]
as $z\rightarrow 0$; thus the logarithm behaves like $\log\epsilon$. As $\epsilon\log\epsilon\rightarrow 0$,
\[\int_{C_{\epsilon}}f(z)\,dz\rightarrow 0\]
Therefore,
\[\frac{1}{\pi}\int_0^{\pi}\log(-2ie^{ix}\sin x)\,dx=0.\]
Now, consider the branch of the logarithm where $\log e^{ix} = ix$. Therefore, $\log(-i)=-\frac{\pi i}{2}$ and thus we see that
\[\frac{1}{\pi}\left[\pi\log 2-\frac{\pi^2 i}{2}+\int_0^{\pi}\log(\sin x)\,dx+\frac{\pi^2 i}{2}\right]=0\]
which implies that
\[\int_0^1\log(\sin\pi x)\,dx=\frac{1}{\pi}\int_0^{\pi}\log(\sin x)\,dx = -\log 2\]
and the proof is complete.
(I think this is good enough...let me know if there are any faulty assumptions on my part... XD)
EDIT #1: The diagram is incorrect (as I noticed when I pulled this from an old homework assignment in my complex analysis course; I'll update it in a few.)
EDIT #2: Updated the contour diagram.
EDIT #3: Fixed some typos.