Edit: I figured out a way to prove the claim. I have attached a solution as my own answer.
Let $f$ be such that $|f(x)|\leq 1, |f'(x)|\leq M$ on a real interval $[a, b]$. I am reading a proof which claims as a subclaim (without proof) that if $A\subset [a, b]$ is a subinterval of length $rM^{-1}$ with $r := M^{1/10}||f||_{L^2([a, b], dx)}^{6/10}$ such that $\sup_{x\in A}|f(x)|\geq 2r$, then $|f(x)|\geq r$ on $A$. This may be an elementary result since I have not managed to find any way to use the assumption on the bound of $|f'(x)|$.
As of now, I have no work to show for this question. However, I will point out that to my intuition this claim feels "wrong" because by assumption only that $|f|$'s maximum value on $A$ (by $f$'s continuity on $[a, b]$) is at least $2r$. If $A$ were to be any subinterval of $[a, b]$, this wouldn't be that useful since $f$ could behave quite wildly on $A$. But I suppose that the gist of the inequality is the fact that $A$'s length is tied to $|f'|$'s maximum value, so $f$, as a continuous function, doesn't have "much room" to change from $2r$ to anything else.
My stated intuition in the form "$f$ doesn't have enough room to change on $A$" turned out to be exactly what I needed. Specifically, write $A = [c, d]\subset [a, b]$ as a subinterval of length $rM^{-1}$ such that $\sup_{x\in [c, d]}|f(x)|\geq 2r$. Since $f$ is continuous on $[a, b]$, it attains its maximum value on $[c, d]$. Suppose that $t$ is one such point that $|f(t)| = \sup_{x\in [c, d]}|f(x)|$. By fundamental theorem of calculus we have that $f(s_2) - f(s_1) = \int_{s_1}^{s_2}f'(x)dx$. Then for any $s\in [c, d]$, we have
$$|f(s)| = |f(s) - f(t) + f(t)|\geq ||f(s) - f(t)| - |f(t)||$$
whence
$$|f(s)| \geq |f(t)| - |f(s) - f(t)|\geq 2r - |f(s) - f(t)|$$
Since $f(s) - f(t) = \int_{t}^{s}f'(x)dx$ we have
$$|f(s) - f(t)|\leq |s - t|M\leq rM^{-1}M = r$$
Hence for any $s\in [c, d]$,
$$|f(s)|\geq 2r - |f(s) - f(t)|\geq 2r - r = r$$
as claimed.