[proofplan] We verify that $X = F^{-1}(U)$ has cdf $F$ by computing $\mathbb{P}(X \le x)$ directly. Since $F$ is strictly increasing and continuous, $F^{-1}$ is well-defined, and the event $\{F^{-1}(U) \le x\}$ is equivalent to $\{U \le F(x)\}$. The result then follows from the cdf of the uniform distribution. [/proofplan] [step:Reduce $\mathbb{P}(F^{-1}(U) \le x)$ to $\mathbb{P}(U \le F(x))$ using monotonicity] Since $F$ is strictly increasing and continuous, $F$ has a well-defined inverse $F^{-1} : (0, 1) \to \mathbb{R}$, which is also strictly increasing. Applying $F$ (which preserves the direction of inequalities) to both sides of $F^{-1}(U) \le x$, \begin{align*} F^{-1}(U) \le x \iff F(F^{-1}(U)) \le F(x) \iff U \le F(x). \end{align*} The first equivalence uses the fact that $F$ is strictly increasing, so $a \le b \iff F(a) \le F(b)$. The second uses $F \circ F^{-1} = \operatorname{id}$ on $(0,1)$. [/step] [step:Evaluate $\mathbb{P}(U \le F(x))$ using the uniform cdf] Since $F$ is a cdf mapping $\mathbb{R}$ to $(0,1)$ (strictly increasing and continuous with $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to \infty} F(x) = 1$), the value $F(x)$ lies in $(0, 1)$ for all $x \in \mathbb{R}$. Since $U \sim U[0, 1]$, \begin{align*} \mathbb{P}(X \le x) = \mathbb{P}(U \le F(x)) = F(x), \end{align*} where the last equality is the cdf of the uniform distribution: $\mathbb{P}(U \le u) = u$ for $u \in [0, 1]$. Therefore $X = F^{-1}(U)$ has cdf $F$. [/step]