Here is the start of a solution using power series. To fill in the arguments it may be helpful to assume that $0 < \delta \leq 1/2$, which is reasonable since the the interval $[\delta,1-\delta]$ is empty for $\delta > 1/2$.
As TMM noted, $\sin(\pi x)$ is symmetric about $x=1/2$ (where it has a local maximum), so for $x \in [\delta,1-\delta]$,
$ \begin{align} \sin(\pi x) &\geq \sin(\pi \delta) \\ &= \pi \delta - \frac{\pi^3\delta^3}{3!} + \frac{\pi^5\delta^5}{5!} - \frac{\pi^7\delta^7}{7!} + \cdots \\ &\geq \pi \delta - \frac{\pi^3 \delta^3}{3!}. \end{align} $
Now, just show that
$ \pi \delta - \frac{\pi^3\delta^3}{3!} - \frac{\pi\delta}{2} > 0 $
to get the result.