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.