As the Wikipedia page states, "Unfortunately, the distribution does not, in general, have an inverse. One may define, for $y \in [0,1]$, the generalized inverse distribution function $F^{-1}(y)= \inf \{x: F(x) \geq y\}$."
Since not all distribution functions $F$ are strictly increasing, not all $F$ have (proper) inverses. When $F$ does has a proper inverse, then the two definitions coincide. When $F$ does not have an inverse (example?) then the definition given still retains many useful properties used in probabilistic arguments.
$F^{-1}$ is simply the notation used in this context to be suggestive of an inverse. In some books, the notation $F^{\leftarrow}$ is used to distinguish it or make clear that a (proper) inverse may not exist and the generalized version is being used. However, this alternate notation is not all that common.
Let's briefly prove some of the facts on the Wikipedia webpage.
Fact 1: $F^{-1}(y)$ is nondecreasing.
Proof: Let $y_1 < y_2$. Then $F(x) \geq y_2$ implies $F(x) \geq y_1$, and so $\{x: F(x) \geq y_1\} \supset \{x:F(x) \geq y_2\}$. Hence, clearly, $\inf \{x: F(x) \geq y_1\} \leq \inf \{x : F(x) \geq y_2\}$.
Fact 2: $F^{-1}(F(x)) \leq x$.
Proof: $F^{-1}(F(x)) = \inf\{z: F(z) \geq F(x)\}$ and $x \in \{z: F(z) \geq F(x)\}$, so the result follows.
Fact 3: $F( F^{-1}(y) ) \geq y$.
Proof: Let $x_n \in \{x : F(x) \geq y\}$ such that $x_n \to x_0$. Then $\liminf_n F(x_n) \geq y$, but since $F$ is monotone nondecreasing and continuous from the right, $\liminf_n F(x_n) \leq F(x_0)$. Hence $F(x_0) \geq y$, that is, $x_0 \in \{x: F(x) \geq y\}$. So, $\{x: F(x) \geq y\}$ is closed and hence contains its infimum. Thus, $F(F^{-1}(y)) \geq y$.
Fact 4: $F^{-1}(y) \leq x$ iff $y \leq F(x)$.
Proof: $F^{-1}(y) \leq x$ implies $x \in \{z: F(z) \geq y\}$ and so $y \leq F(x)$. On the other hand, if $y \leq F(x)$ then $x \in \{z : F(z) \geq y\}$ and so $F^{-1}(y) \leq x$ since $F^{-1}(y)$ is the infimum.
Best Answer
Let $U$ be uniformly distributed on $[0,1]$ and let $\bar X:= Q(U)$.
Composing by $Q\circ F$ on both sides, it holds that : $$ Q(F(\bar X)) = Q(F(Q(U))\;\;\text{ almost surely}$$
To make it slightly easier to read, I rewrite it as follows : $$ Q(F(\bar X)) = Q(F\circ Q(U)) \;\;\text{ almost surely} \tag1$$
Now, by definition of $Q$, it holds that for all $p\in\mathbb R$, $F\circ Q(p)\ge p$. Applying $Q$ (which is non-decreasing) on both sides gives $Q(F\circ Q(p))\ge Q(p) $ for all $p$, from which we deduce that :
$$ Q(F\circ Q(U)) \ge Q(U) = \bar X\;\;\text{ almost surely} \tag A$$
Now, again by definition of $Q$, we have that for all $p\in\mathbb R$, $Q(F(p)) \le p $, which implies that : $$Q(F(\bar X)) \le \bar X\;\;\text{ almost surely} \tag B$$
From $(A)$ and $(B)$ we deduce that $$Q(F(\bar X)) =\bar X\;\;\text{ almost surely} $$
To conclude, note that $\bar X$ and $X$ have the same distribution (I let you do the proof), so for any set $S$, it holds that $\mathbb P(X\in S) = \mathbb P(\bar X\in S)$. In particular if we consider the set $S:=\{x\in \mathbb R \ |\ Q(F(x)) =x\}$, we have that $$ \mathbb P(\{Q(F(X) = X\}) =\mathbb P(X\in S) = \mathbb P(\bar X\in S) =\mathbb P(\{Q(F(\bar X) =\bar X\}) = 1 $$
And we are done.