[proofplan]
We first verify that the infimum defining $Q(p)$ is finite for every $p\in(0,1)$, using the two tail limits in the definition of a distribution function. Monotonicity follows from the nesting of the superlevel sets $\{x:F(x)\ge p\}$. For left-continuity, we compare $Q(p)$ with the supremum of the values $Q(q)$ for $q<p$; if that supremum were strictly below $Q(p)$, a point between them would force a contradiction by the defining infimum property of $Q(q)$.
[/proofplan]
[step:Verify that each quantile is a finite real number]
For $r\in(0,1)$, define the superlevel set $A_r\subset\mathbb R$ by
\begin{align*}
A_r:=\{x\in\mathbb R:F(x)\ge r\}.
\end{align*}
Since $F$ is a distribution function, it is nondecreasing and satisfies
\begin{align*}
\lim_{x\to-\infty}F(x)=0
\end{align*}
and
\begin{align*}
\lim_{x\to\infty}F(x)=1.
\end{align*}
Because $r<1$, there exists $b\in\mathbb R$ such that $F(b)\ge r$, so $A_r$ is nonempty. Because $r>0$, there exists $a\in\mathbb R$ such that $F(a)<r$. If $y\le a$, then monotonicity of $F$ gives $F(y)\le F(a)<r$, so $y\notin A_r$. Hence $a$ is a lower bound for $A_r$.
Thus $A_r$ is nonempty and bounded below, so its infimum is a real number. Therefore $Q(r)=\inf A_r\in\mathbb R$ for every $r\in(0,1)$.
[/step]
[step:Prove monotonicity by nesting the defining superlevel sets]
Let $r,s\in(0,1)$ satisfy $r\le s$. With $A_r$ and $A_s$ as defined above, if $x\in A_s$, then $F(x)\ge s\ge r$, so $x\in A_r$. Hence
\begin{align*}
A_s\subset A_r.
\end{align*}
Since both sets are nonempty and bounded below, inclusion reverses their infima:
\begin{align*}
\inf A_r\le \inf A_s.
\end{align*}
Therefore
\begin{align*}
Q(r)\le Q(s).
\end{align*}
This proves that $Q$ is nondecreasing on $(0,1)$.
[guided]
Fix $r,s\in(0,1)$ with $r\le s$. The quantile $Q(r)$ is defined from the set of points where $F$ has reached level $r$, while $Q(s)$ is defined from the set of points where $F$ has reached the higher level $s$. Formally, set
\begin{align*}
A_r:=\{x\in\mathbb R:F(x)\ge r\}
\end{align*}
and
\begin{align*}
A_s:=\{x\in\mathbb R:F(x)\ge s\}.
\end{align*}
If $x\in A_s$, then $F(x)\ge s$. Since $s\ge r$, this implies $F(x)\ge r$, and hence $x\in A_r$. Thus
\begin{align*}
A_s\subset A_r.
\end{align*}
The direction of the inequality between infima is important. The larger set can begin no later than the smaller set. Since $A_r$ contains $A_s$, every lower bound for $A_r$ is also a lower bound for $A_s$, and equivalently the greatest lower bound of $A_r$ cannot exceed the greatest lower bound of $A_s$. Hence
\begin{align*}
\inf A_r\le \inf A_s.
\end{align*}
Using the definition of $Q$, this is exactly
\begin{align*}
Q(r)\le Q(s).
\end{align*}
Since the argument holds for every $r,s\in(0,1)$ with $r\le s$, the function $Q:(0,1)\to\mathbb R$ is nondecreasing.
[/guided]
[/step]
[step:Compare $Q(p)$ with the supremum of the left-hand quantiles]
Fix $p\in(0,1)$. Define the index set $I_p\subset(0,1)$ by
\begin{align*}
I_p:=\{q\in(0,1):q<p\}.
\end{align*}
Since $p>0$, the number $p/2$ belongs to $I_p$, so $I_p$ is nonempty. Define the set of left-hand quantile values $B_p\subset\mathbb R$ by
\begin{align*}
B_p:=\{Q(q):q\in I_p\}.
\end{align*}
By monotonicity, for every $q\in I_p$,
\begin{align*}
Q(q)\le Q(p).
\end{align*}
Thus $B_p$ is nonempty and bounded above. Define
\begin{align*}
L:=\sup B_p.
\end{align*}
Then
\begin{align*}
L\le Q(p).
\end{align*}
[/step]
[step:Rule out a strict gap between the left supremum and $Q(p)$]
Assume, for contradiction, that
\begin{align*}
L<Q(p).
\end{align*}
Choose $x\in\mathbb R$ such that
\begin{align*}
L<x<Q(p).
\end{align*}
Since $x<Q(p)=\inf A_p$, where $A_p:=\{y\in\mathbb R:F(y)\ge p\}$, the point $x$ cannot belong to $A_p$. Hence
\begin{align*}
F(x)<p.
\end{align*}
Choose $q\in(0,1)$ such that
\begin{align*}
F(x)<q<p.
\end{align*}
Then $q\in I_p$. We claim that $x\le Q(q)$. Indeed, if $y\in A_q:=\{z\in\mathbb R:F(z)\ge q\}$ and $y<x$, then monotonicity of $F$ gives
\begin{align*}
F(y)\le F(x)<q,
\end{align*}
contradicting $y\in A_q$. Therefore every element of $A_q$ is at least $x$, so $x$ is a lower bound for $A_q$. Hence
\begin{align*}
x\le \inf A_q=Q(q).
\end{align*}
But $q\in I_p$, so $Q(q)\in B_p$, and therefore $Q(q)\le L$. Combining the inequalities gives
\begin{align*}
x\le Q(q)\le L<x,
\end{align*}
a contradiction. Thus $L=Q(p)$.
[/step]
[step:Identify the left-hand limit with the supremum]
Let
\begin{align*}
S:=\sup\{Q(q):q\in(0,1),q<p\}.
\end{align*}
The preceding step gives $S=Q(p)$. We verify directly that $S$ is the left-hand limit. Let $\varepsilon>0$. By the definition of supremum, there exists $q_0\in(0,1)$ with $q_0<p$ such that
\begin{align*}
S-\varepsilon<Q(q_0)\le S.
\end{align*}
For every $q\in(q_0,p)$, monotonicity of $Q$ gives
\begin{align*}
S-\varepsilon<Q(q_0)\le Q(q)\le S.
\end{align*}
Thus $Q(q)\to S$ as $q\uparrow p$. Since $S=Q(p)$, we obtain
\begin{align*}
\lim_{q\uparrow p}Q(q)=Q(p).
\end{align*}
Since $p\in(0,1)$ was arbitrary, $Q$ is left-continuous on $(0,1)$. Together with monotonicity, this proves the theorem.
[/step]
