Question: Let $X$ be a random variable such that it's distribution function $F(x):=P(X \le x)$ is continuous. 

Then is it true that $F(X)$ follows a uniform distribution ?

I can show this if $F$ is differentiable, but otherwise I'm not sure. 

Please help.

Answer: 
Yes, it follows a uniform distribution.

We know that $F$ is non-decreasing, and we can further note that 

$$P(F(X) \leq x) = P(X \in \{y \mid F(y) \leq x\})$$

Now set $M = \sup\{y \mid F(y) \leq x\} $ so that by continuity, we can obtain $F(M) = x$. It follows that $P(F(X) \leq x) = x$ for $0\leq x \leq 1$. In other words, $F(X)$ is uniformly distributed.
For convenience, consider that $F$ is absolutely continuous and strictly monotone. Then the random variable $U = F\_{X}(x)$ has distribution given by
\begin{align\*}
G\_{U}(u) = \textbf{P}(U\leq u) = \textbf{P}(F\_{X}(x)\leq u) = \textbf{P}(x \leq F^{-1}\_{X}(u)) = F\_{X}(F^{-1}\_{X}(u)) = u
\end{align\*}

from whence we conclude $U = F\_{X}(x)$ has uniform distribution.