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Let $y\in(0,1)$. Since $F$ is continuous, there exists $x\in\mathbb{R}$ s.t. $F(x)=y$. Thus, $$ \mathsf{P}(Y\le y)=\mathsf{P}(F(X)\le F(x))=F(x)=y, $$ i.e., $Y\sim\text{U}[0,1]$. In order to see the first equality we don't need continuity. Specifically, since any cdf is right-continuous, \begin{align} \{F(X)\le F(x)\}&=\{\{F(X)\le F(x)\}\cap\{X\le x\}\}\cup \{\{F(X)\le F(x)\}\cap\{X>x\}\} \\ &=\{X\le x\}\cup \{\{F(X)=F(x)\}\cap\{X>x\}\}, \end{align} and $\mathsf{P}(\{F(X)=F(x)\}\cap\{X>x\})=0$.

Let $y\in(0,1)$. Since $F$ is continuous, there exists $x\in\mathbb{R}$ s.t. $F(x)=y$. Thus, $$ \mathsf{P}(Y\le y)=\mathsf{P}(F(X)\le F(x))=F(x)=y, $$ i.e., $Y\sim\text{U}[0,1]$. In order to see the first equality we don't need continuity. Specifically, since any cdf is right-continuous, \begin{align} \{F(X)\le F(x)\}&=\{\{F(X)\le F(x)\}\cap\{X\le x\}\}\cup \{\{F(X)\le F(x)\}\cap\{X>x\}\} \\ &=\{X\le x\}\cup \emptyset. \end{align}

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