We are given:
\[
-1 \leq \sin x \leq 1
\]
Now,
\[
1 - \sin x \geq 0 \quad \text{(since \( \sin x \leq 1 \))}
\]
\[
1 + 1 \geq \sin x + 1 \geq 1 - 1
\]
\[
0 \leq 1 - \sin x \leq 2
\]
Thus, the minimum value of \( 1 - \sin x \) is \( 0 \), which corresponds to option (4).