There are two mutually inverse facts.
If we take a r.v. $X$ with C.D.F. $F(x)$, and $F(x)$ is continuous and strictly increasing, then r.v. $Y=F(X)$ is uniformly distributed over $(0,1)$.
In fact, "strictly increasing" for this statement is not required. But let it be.
And the inverse:
If we take a r.v. $Y$ which is uniformly distributed over $(0,1)$ and some C.D.F. $F(x)$ which is continuous and strictly increasing, then r.v. $X=F^{-1}(Y)$ has C.D.F. $F(x)=\mathbb P(X\leq x)$.
We have continuous and strictly increasing function $F$ which satisfies all the requirements to be a C.D.F. of some random variable. Namely, it has zero limit at $-\infty$ and unit limit at $+\infty$.
We want to construct a r.v. $X$ such that $F$ is the C.D.F. of $X$.
To do it, one need to take $Y$ uniformly distributed on $(0,1)$ and $X=F^{-1}(Y)$. The last equality is equivalent to the equality $Y=F(X)$ since $F$ is invertible.
Both facts are checked simply. Take the second one: let $Y$ is uniformly distributed over $(0,1)$ and $F(x)$ is continuous and strictly increasing C.D.F. Define $X=F^{-1}(Y)$. Find C.D.F. of $X$:
$$
F_X(x)=\mathbb P(X\leq x) = \mathbb P(F^{-1}(Y)\leq x) = \mathbb P(Y\leq F(x)) = F(x).
$$
since $\mathbb P(Y\leq a)=a$ for $0<a<1$. We conclude that $F(x)$ equals to C.D.F. of r.v. $X=F^{-1}(Y)$.
As an example, if $Y$ is uniformly distributed over $(0,1)$, then
- $X=-\ln(1-Y)$ is exponentially distributed with $F_X(x)=1-e^{-x}$ ($x>0$). Another words, if $X$ is such that $$Y=1-e^{-X},$$
then $X$ has C.D.F. $F_X(x)=1-e^{-x}$ ($x>0$).
- If $\Phi(x)$ is C.D.F. of standard normal distribution then $X$ such that $$Y=\Phi(X)$$ has standard normal distribution. One can write $X$ as $X=\Phi^{-1}(Y)$.
And so on.
So, the answer to your question in the header is "Yes". If proper function of $X$ is uniformly distributed, then this function is C.D.F. of $X$.
If you are allowed to assume $P(X=x)=0$, you're proof is fine, but how would you proof $P(X=x)=0$? Actually you are trying to proof that the CDF has no jumps, and you use the fact that the CDF has no jumps.
I think it would be best to note that a random variable is continuous when it has a probability density function $f$, and we can write $P(X\leq x)=\int_{-\infty}^xf(u)\mathrm{d}u$. Continuity of the continuous random variable now follows directly from the fact the Riemann integrals are continuous.
If you want to proof that $P(X=x)=0$: $$P(X=x)=\lim_{\epsilon\to0^+}P(x-\epsilon<X\leq x)=\lim_{\epsilon\to0^+}[F_X(x)-F_X(x-\epsilon)]=\lim_{\epsilon\to0^+}\int_{x-\epsilon}^xf_X(u)\mathrm{d}u=0,$$ where we use the continuity of the probability measure in the first equality.
Best Answer
Let $F_Y(y)$ be the CDF of $Y = F(X)$. Then, for any $y \in [0,1]$ we have:
$F_Y(y) = \Pr[Y \le y] = \Pr[F(X) \le y] = \Pr[X \le F^{-1}(y)] = F(F^{-1}(y)) = y$.
What distribution has this CDF?