Show that if the inverse cumulative distribution function is continuous, it is invertible Let $F:\mathbb R\to[0,1]$ be a distribution function$^1$ and note that $$F^{-1}(t):=\inf\left\{x\in\mathbb R:F(x)\ge t\right\}\;\;\;\text{for }t\in(0,1)$$ is left-continuous and nondecreasing with $$F^{-1}(t)\le x\Leftrightarrow t\le F(x)\;\;\;\text{for all }x\in\mathbb R\text{ and }t\in(0,1)\tag1.$$

Question 1: Let $t\in(0,1)$. How can we show that if $F$ is continuous at $x:=F^{-1}(t)$, then $F(x)=t$?
Question 2: Now let $F_1,F_2$ be distributions functions, $s\in(0,1)$ and $x:=F_1^{-1}(s)$. Assume $x\in F_2^{-1}((0,1))$. I want to determine a $t\in(0,1)$ such that $F_2^{-1}(t)=x$. I guess we need to distinguish whether $F_2$ is continuous at $x$ (in which case there should be a unique solution) or not (in which case there should be an solution interval).

I guess both questions are easy to prove, but somehow I've got problems to solve them. Regarding the first question: I've tried to assume the contrary, i.e. $\varepsilon:=|F(x)-t|>0$, but I wasn't able to derive a contradiction from that.
Regarding the second question: In the first question we've shown that if $(F\circ F^{-1})(t)=t$ for all $t\in(0,1)$ such that $F$ is continuous at $F^{-1}(t)$. Now I guess that we need to show something similar: If $F$ is continuous at $x\in\mathbb R$, then $(F^{-1}\circ F)(x)=x$.
With this result we would see that if $F_2$ is contiunous at $x:=F_1^{-1}(s)$, then we may choose $t=(F_2\circ F_1^{-1})(s)=F_2(x)$.

$^1$ i.e. $F$ is right-continuous and nondecreasing with $F(-\infty):=\lim_{x\to-\infty}F(x)=0$ and $F(\infty):=\lim_{x\to\infty}F(x)=1$.
 A: Question 1: Let $x_0 = F^{-1}(t)$. Since $F$ is continuous at $x_0$, $$\lim_{n \to \infty} F(x_0 - n^{-1}) = F(x_0) = \lim_{n \to \infty} F(x_0 + n^{-1})$$
Since $x_0 - n^{-1} < x_0$ we have that $F(x_0 - n^{-1}) < t$ so that $\lim_{n \to \infty} F(x_0 - n^{-1}) \leq t$. Also, by definition of $x_0$, for every $n$ there is an $x_1 < x_0 + n^{-1}$ such that $F(x_1) \geq t$. Since $F$ is non-decreasing we have that $F(x_0 + n^{-1}) \geq t$ and so $\lim_{n \to \infty} F(x_0 + n^{-1}) \geq t$. Combining this we have that $t \leq F(x_0) \leq t$ which implies the desired result.
Question 2: [This part of the answer was written before the assumption $x \in F_2^{-1}((0,1))$ was added, which invalidates it]
This result is false (assuming the $F$ on the right hand side is meant to be one of $F_1$ or $F_2$). For example let $F_1$ be the distribution function of a uniform random variable on $[0,1]$ and $F_2$ be the distribution function of the constant random variable $0$. Then for any $t \in (0,1)$, we have that $F_2^{-1}(t) = 0$. However, e.g. $F_1^{-1}(\frac12) = \frac12$ so we can set $x = \frac12$. We then have that
$$\{ t \in (0,1): F_2^{-1}(t) = x\} = \emptyset \neq \{1\} = \{F_2(\frac12)\}   = [F_2(\frac12 -), F_2(\frac12+)]$$
We also have that $\emptyset \neq \{F_1(\frac12)\} = \{\frac12\}$. 

For the current formulation of question $2$, you can always take $t = F_2(x)$. Indeed, if $y \geq x$ then $F_2(y) \geq F_2(x) = t$ so that $F_2^{-1}(t) \leq x$. If $F_2^{-1}(t) < x$ then there is $x_1 < x$ such that $F_2(x_1) \geq t = F_2(x)$. Since $F_2$ is non-decreasing this means that $F_2$ is constant on $[x_1,x]$. But then $x \not \in F_2^{-1}((0,1))$ since if $x \in \{y : F_2(y) \geq t_0\}$ then $x_1 \in \{y: F_2(y) \geq t_0\}$ also.
A: Question 1: Let $x^+_n$ be a sequence approaching $x$ from above, (and $x^-_n$ approach from below). We will make none of them equal to $x$; just that the limits of these two sequences are $x$. It is easy to see that $F(x^+_n) \geq t$, since by definition, $x$ is a lower bound for all $x_i$s that satisfy that property, and $F$ is monotone non-decreasing. Similarly, we have that $F(x^-_n) \leq t$, since if for any $x^-_n < x$ was greater than $t$, $x$ could not be the infimum (we are invoking monotonicity here again). Taking limits now gives you what you want.
Question 2: (I don't really understand why you have $F_1$ and $F_2$ there; I am taking them both equal to $F$) If the upper and lower limits are different, you actually need to make the lower limit an open bracket: Take $F$ to be a half step at 0 and another half step at 1, and let $x$ equal to $1$. The limit of $F(x)$ approaching $1$ from the bottom will be 1/2, but $F^{-1}(1/2)$ is not 1. Besides that I think the above proof mostly goes through.  
Note: I was actually writing a proof claiming the claim was correct; I stopped after getting stuck only to realize the claim was incorrect. I am leaving it behind as a potential reference to writing out the proof of the second claim (might get back to it later if there is demand, or if someone wants to pick it up from there on)
 Question 2: Yes to both claims (though I don't really understand why    you have $F_1$ and $F_2$ there; I am taking them both equal to $F$).    We need to show that the two sets contain each other. Let $t$ be such    that $F^{-1}(t) = x$. The proof is essentially the same as last part,    where you get $F(x^+_n) \geq t$ and $F(x^-_n) \leq t$, so $t$ is    between those limits. 
       Now, just run the arguments backwards; i.e. conversely, take a $t$    between those limits, then $t$ satisfies those two individual    inequalities. Now, we invoke the fact that $x$ is indeed the inverse    of some $s$, (the statement would fail in flat regions of $F$    otherwise). 
       Note that $x$ is the best lower bound for $F^{-1}(s)$, and so is also    the best lower bound for all $t$ in the interval. Suppose not.    Suppose that there is a different $x'$ for some other $t$ in the    interval; this $x'$ is either bigger or smaller than $x$. If $t > s$,    then $x' > x$, however, we know that there exists a sequence $x^+_n$    that gets arbitrarily close to $x$ that satisfies $F(x^+_n) \geq t$,    so $x'$ cannot be the lower bound. 
       The other case $t < s$ is surprisingly more complicated.
