Let $F:\mathbb R\to[0,1]$ be a distribution function$^1$ and $$M_t:=\left\{x\in\mathbb R:F(x)\ge t\right\}$$ and $$F^{-1}(t):=\inf M_t$$ for $t\in[0,1]$.
If $t>0$, can we show that $x_0:=F^{-1}(t)\in\mathbb R$ and $F(x_0)\ge t$?
I'm not sure how to argue, but assume for the moment that we already know that $x_0\in\mathbb R$. Assume that $F(x_0)<t$ so that $\varepsilon:=t-F(x_0)>0$. Since $F$ is right-continuous at $x_0$, there is a $\delta>0$ with $$|F(x)-F(x_0)|<\varepsilon\;\;\;\text{for all }x\in(x_0,x_0+\delta)\tag1$$ and hence $$F(x)\le F(x_0)+|F(x)-F(x_0)|<t\;\;\;\text{for all }x\in(x_0,x_0+\delta)\tag2.$$ Thus, $$M_t\subseteq\mathbb R\setminus(x_0,x_0+\delta)=(-\infty,x_0]\cup[x_0+\delta,\infty)\tag3.$$ Since $x_0$ is a lower bound for $M_t$ and, by assumption, $x_0\not\in M$, $$M_t\subseteq[x_0+\delta,\infty)\tag4.$$ Is this a contradiction to the definition of $x_0$?
$^1$ i.e. $F$ is right-continuous and nondecreasing with $F(-\infty):=\lim_{x\to-\infty}F(x)=0$ and $F(\infty):=\lim_{x\to\infty}F(x)=1$.
I think I know what question you meant to ask, but you wrote it in a confusing (more precisely, incorrect) way since $M_t$ is treated as both a set and a number in your question. (It's a set in the definition you wrote and in your equations $(3)$ and $(4)$, but it's a number in the highlighted portion of your question.)
So here is what I think your question is a special case of:
Note that $F$ doesn't need to be a distribution function, and $t$ does not have to positive.
Why is this true? Well, if we write $y$ for the infimum of the set of $x\in\mathbb R$ satisfying $F(x)\geq t$ then we can find a sequence $x_1,x_2,\ldots$ tending to $y$ from above such that $F(x_n)\in [t,\infty)$ for all $n$. By right continuity, we have that $F(x_n)$ tends to $F(y)$, and thus since $[t,\infty)$ is closed it follows that $F(y)\in [t,\infty)$ as well, proving the inequality.