For any given $F_X$, show that if $U \sim \text{Uniform}(0,1)$ and $F_{X}^{-}(u) = \text{min}\{x:F_X(x)\geq u\}$ then $X = F_{X}^{-}(U)$ has distribution $F_X$.
I tried this and got to $F_X(x) = P(X\leq x) = P(F_{X}^{-}(U)\leq x)$
Not really sure how to proceed. Also, $F_X$ is not necessarily continuous.
Although the CDF can have jumps, it's still helpful to make some sketches where it IS continuous and realize that $F_X^{-}$ in that case is exactly the inverse mapping of $F_X$ (the curve is flipped with respect to the diagonal line). Thus we know this is essentially that well-known transformation in disguise.
Now formally, just plugin the definition (allow me to alter the notations to make the distinction explicit) $$ \Pr(X \leq t ) = \Pr\left( F_{X}^{-}(U)\leq t \right) = \Pr\bigl( \min\left\{y: F_{X}(y) \geq U \right\} \leq t \bigr) \label{Eq_1} \tag*{Eq.(1)} $$ which literally translates to "$t$ is larger than such $y$", where we have $y$ satisfying $F_{X}(y) \geq U$. With such $y$ being the minimal among those values, we have $$ \left. \begin{aligned} F_{X}(y) &\geq U \\ t &\geq y \end{aligned} \right\} \implies F_{X}(t) \geq U ~,\qquad \text{or} \quad U \leq F_{X}(t)$$ because by definition any CDF is non-decreasing (including discontinuous ones).
Therefore, continuing \ref{Eq_1} \begin{align} \Pr(X \leq t ) &= \Pr\bigl( \min\left\{y: F_{X}(y) \geq U \right\} \leq t \bigr) \\ &= \Pr\bigl( U \leq F_{X}(t) \bigr) \\ &= F_{X}(t) \end{align} because $U$ has a uniform distribution.
So we have proved that $X$ has a CDF of $F_X$. I hope this argument is rigorous enough for you.