$f:[0,1]\rightarrow [\underline{f},\overline{f}]$ is continuous, "does not oscillate", i.e. $|\{f(x)=v\}|<\infty$ for all $v \in [\underline{f},\overline{f}]$, has a unique global minimum at $x^* \in (0,1)$. Must it be monotonically decreasing in $(x^*-\epsilon, x^*)$ for some $\epsilon>0$?
If not, for all $\epsilon>0$, $\exists \;x' \in(x^*-\epsilon, x^*)$ such that $x'$ is a local maxima of $f$. But does that violate non-oscillation necessarily? I'm not able to figure out.
It doesn't. Take, for example, $f(x) = x^2 \cdot \left(2 + \sin\left(\frac{1}{x}\right)\right)$ (it's on $[-1, 1]$, not $[0, 1]$, but it obviously doesn't matter).
This function has global minimum at $x = 0$ (as $f(0) = 0$ and $f(x) > 0$ if $x \neq 0$). It's not oscillating by your definition: it's zero only in one point, and any value $\epsilon > 0$ it achieves only if $x > \sqrt{\epsilon / 3}$, where everything is clearly fine ($\sin$ changes sign only finitely many times).