Let $F:\mathbb R\to[0,1]$ be nondecreasing and right-continuous and $$F^{-1}(u):=\inf\underbrace{\{x\in\mathbb R:F(x)\ge u\}}_{=:\:I_u}\;\;\;\text{for }u\in[0,1].$$
Question 1: Let $x\in\mathbb R$ and $u\in[0,1]$. I want to show that $$F^{-1}(u)\le x\Leftrightarrow u\le F(x)\tag1.$$
The crucial thing in this setting is that we don't restrict $F^{-1}$ to $(0,1)$ and we don't assume $\lim_{x\to\infty}F(x)=1$. So, $F^{-1}(0)=-\infty$ and if $\sup_{\mathbb R}F<1$, then $F(1)=\infty$. However, this should not be a problem or am I missing something?
In fact, the direction "$\Leftarrow$" is obviously true by definition of the infimum. For the other direction, assume $u>F(x)$. Since $F$ is nondecreasing, $$u>F(y)\;\;\;\text{for all }y\le x\tag2.$$ And since $F$ is right-continuous at $x$, there is a $\delta>0$ with $$F(y)<u\;\;\;\text{for all }y\in[x,x+\delta)\tag3.$$ So, $$I_u\subseteq[x+\delta,\infty)$$ and hence $$F^{-1}(u)\ge x+\delta>x\tag4.$$
Question 2: By $(1)$, $$\{u\in[0,1]:F^{-1}(u)\le x\}=[0,F(x)]\tag5$$ and hence $F^{-1}$ is Borel measurable. Aren't we immediately able to conclude that $$X:=F^{-1}$$ is an $\overline{\mathbb R}$-valued random variable on $([0,1],\mathcal B([0,1]),\mathcal U_{[0,\:1]})$, where $\mathcal U_{[0,\:1]}$ denotes the uniform distribution on $[0,1]$, with $$\mathcal U_{[0,\:1]}[X\le x]=F(x)\;\;\;\text{for all x}\in\mathbb R,\tag6$$ which is to say that $F$ is the distribution function of $X$?
I agree with what you wrote.
I'd add that $\lim_{x\to+\infty}F(x)=\mathcal U_{[0,1]}(X<+\infty)$.
And you didn't mention the limit at $-\infty$ but we also have $\lim_{x\to-\infty}F(x)=\mathcal U_{[0,1]}(X=-\infty)$.