Let 
\begin{equation}
	N:=\inf\{n\ge2\colon X_{n-1}>X_n\}. 
\end{equation}
We want to find 
\begin{equation}
	EX_N=\sum_{n=2}^\infty EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n).
\end{equation}
Next, 
\begin{equation}
\begin{aligned}
	&EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n) \\
	&=EX_n\,1(X_1\le\cdots\le X_{n-1})-EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n), 
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
	&EX_n\,1(X_1\le\cdots\le X_{n-1}) \\
	&=EX_n\,P(X_1\le\cdots\le X_{n-1})=\frac12\,\frac1{(n-1)!},
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
	&E(1-X_n)\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\ 
	&=E\int_0^1 dx\,1(X_1\le\cdots\le X_{n-1}\le X_n\le x) \\ 
	&=\int_0^1 dx\,P(X_1\le\cdots\le X_{n-1}\le X_n\le x) \\ 
	&=\int_0^1 dx\,x^n\frac1{n!}= \frac1{(n+1)!},
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
	&EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\ 
	&=P(X_1\le\cdots\le X_{n-1}\le X_n)-\frac1{(n+1)!} \\ 
	&=\frac1{n!}-\frac1{(n+1)!},
\end{aligned}	
\end{equation}
\begin{equation}
	EX_N=\sum_{n=2}^\infty \Big(\frac12\,\frac1{(n-1)!}-\frac1{n!}+\frac1{(n+1)!}\Big)=\frac e2-1.
\end{equation}