Most of the time, either the few usual tricks work, or the conjecture is true. In the non-autonomous case, unfortunately, the answer is the former. This is the first counterexample I ever learnt to uniqueness (and I bet it's true for a lot of people):
$$ x' = 2 max(x,0)^{1/2} $$
That has two solutions, if $x(0) = 0$. One is $x(t) = 0$ and one $x(t) = t^2\cdot 1_{t>0}$. Because you want to assume boundedness (as you will see it's not really important, let's consider the bounded counterpart
$$ x' = 2\cdot \begin{cases} 0 & \text{if }t\le 0 \\ \sqrt{x} & \text{if} t\in [0,1]\\ 1 & \text{if }t\ge 1 \\ \end{cases} $$
Let $x(t)$ be a solution to the problem above, and choose an $v>0$. What equation does $z(t) = x(t)+vt$ solve? A direct computation shows:
$$ z' = 2\cdot \begin{cases} v & \text{if }z-vt\le 0 \\ v + \sqrt{z -vt} & \text{if} z-vt\in [0,1]\\ 1+v & \text{if }z-vt\ge 1 \\ \end{cases} $$
So you have a counterexample for nonautonomous solutions. From a philosophical perspective this boils down to the following:
In the non-autonomous case, you have time and space reparametrization symmetries. (At least you reparametrize by smooth changes of coordinates). If you want to show that a statement holds for a certain class of objects, you should look for a class that is invariant under reparametrizations.
Now, in the autonomous case, the class you have given is invariant by the (reduced set of) symmetries of your problem, at least locally, and that gives hope. I only give a sketch of a low-tech proof, comment if something is unclear!
Without loss of generality, $x' = f(x)$, with $M>f(x)>1$ and we want to show existence for $|t|\le 1$ (then you can patch it up). We will assume $f$ is Riemann integrable for simplicity.
Choose a fine partition (in the Riemann integral sense), and let $f^+$, $f^-$ be the two piecewise-constant (upper and lower) functions induced by this partition. With a bit of work, you can show that any solutions to $x'=f(x)$ have to be sandwiched by the unique (uniqueness is tedious but works) solutions to $x'=f^+(x)$ and $x'=f^-(x)$, and that the solutions converge to each other.
Use monotonous convergence theorem to show everything is well defined in the limit.