Here's the problem:
Let $f:[-\frac{\pi}{2},\frac{\pi}{2}] \to [-1,1]$ be the map defined as $f(x) = \sin(x)$ for any $x \in I = [-\frac{\pi}{2},\frac{\pi}{2}]$. Prove that $f$ is surjective.
Proof Attempt:
We know that $f$ is continuous on $\mathbb{R}$ and, therefore, is continuous in the given closed interval. We also know that continuous functions map closed intervals to closed intervals.
In particular, the endpoints of $f(I)$ have to be the absolute extrema of $f$. Now, since $f$ is increasing in $I$, it follows that $min(f(I)) = -1$ and $min(f(I)) = 1$. So, $f(I) = [-1,1]$. This proves the desired result.
Does this proof work? If it doesn't, why? How can I fix it?
"In particular, the endpoints of f(I) have to be the absolute extrema of f".This is not true, imagine if the interval was $[-\pi, \pi]$. What you do know is $\sin(\pi/2) = 1$ and $\sin(-\pi/2) = -1$. Since $\sin$ is continuous, we have by the mean value theorem that all values between -1 and 1 are hit, hence it is surjective.