As I commented under several posts above, the key of a rigorous (and succinct) proof to the general continuous $F$ (that is, $F$ is not necessarily strictly increasing) is by introducing the quantile function:
\begin{align*}
\varphi(u) = \inf\{x: F(x) \geq u\}, \quad 0 < u < 1. \tag{1}\label{1}
\end{align*}
Note that this function is well defined for any distribution function $F$, regardless it has discontinuities or strictly increasing. This can also be used as a definition of the population $u$-quantile of a random variable whose distribution function is $F$. A graph of $\varphi(\cdot)$, which stresses its values at a discontinuity and a plateau point, is shown as follows (source: Probability and Measure (3rd edition), p.189):
The most important property of $\varphi$ is its relationship with $F$: for any $0 < u < 1$:
\begin{align*}
F(\varphi(u)-) \leq u \leq F(\varphi(u)), \tag{2}\label{2}
\end{align*}
where $F(\varphi(u)-) = \lim\limits_{x \uparrow \varphi(u)}F(x)$. From the graph, inequality $\eqref{2}$ is evident. The proof of $\eqref{2}$ will be relegated to the end of this answer. Note that for a general $F$, $\eqref{2}$ is not equivalent to $F(\varphi(u)) = u$, as many "proofs" falsely claimed or relied on.
However, under the condition that $F$ is continuous (everywhere), $\eqref{2}$ of course reduces to $F(\varphi(u)) \leq u \leq F(\varphi(u))$, whence $F(\varphi(u)) = F(\varphi(u)-) = u$. It then follows that
\begin{align*}
P[F(X) < u] &= 1 - P[F(X) \geq u] \\
&= 1 - P[X \geq \varphi(u)] =
P[X < \varphi(u)] = F(\varphi(u)-) = u. \tag{3}\label{3}
\end{align*}
In the second equality above, we used the property that $\{F(X) \geq u\} = \{X \geq \varphi(u)\}$, whose proof is also placed at the end of the answer. Now by
$\eqref{3}$ and the continuity (from above) of the probability measure, we have for $0 < u < 1$:
\begin{align*}
P[F(X) \leq u] = \lim\limits_{h \downarrow 0}P[F(X) < u + h] =
\lim\limits_{h \downarrow 0}(u + h) = u.
\end{align*}
This shows that $F(X) \sim U(0, 1)$, and the proof is complete.
Proof of $\eqref{2}$
By definition $\eqref{1}$, if $y < \varphi(u)$, then $F(y) < u$ (otherwise, $\varphi(u)$ would not be a lower bound of the set $S_u := \{x: F(x) \geq u\}$ as $F(y) \geq u$ and $y < \varphi(u)$). This implies that $F(\varphi(u)-) = \lim\limits_{y \uparrow \varphi(u)}F(y) \leq u$.
On the other hand, for every $n \in \mathbb{N}$, since $\varphi(u) + n^{-1}$ is not the infimum of $S_u$, there exists $x_n \in S_u$ such that $x_n < \varphi(u) + n^{-1}$, whence $F(\varphi(u) + n^{-1}) \geq F(x_n) \geq u$ by the monotonicity of $F$. This implies by the right-continuity of $F$ that $F(\varphi(u)) = \lim\limits_{n \to \infty}F(\varphi(u) + n^{-1}) \geq u$.
This completes the proof of $\eqref{2}$.
Proof of $\{F(X) \geq u\} = \{X \geq \varphi(u)\}$
If $F(X) \geq u$, then $X \in \{x: F(x) \geq u\}$, whence $X \geq \inf\{x: F(x) \geq u\} = \varphi(u)$. This shows that $\{F(X) \geq u\} \subseteq \{X \geq \varphi(u)\}$.
On the other hand, if $X \geq \varphi(u)$, then $F(X) \geq F(\varphi(u))$ by the monotonicity of $F$. As we have proven $F(\varphi(u)) \geq u$, it follows that $F(X) \geq u$, hence $\{X \geq \varphi(u)\} \subseteq \{F(X) \geq u\}$.
