Prove $xe^x =2$ for some $x \in (0,1)$

If $f(x)=2$ some point in the interval $[0,1]$, and we know for a fact that it's not $0$ or $1$, we know that $f(x)=2$ in the interval $(0,1)$.

I post this answer in order to clarify a discusion that arose on Avid's answer.

Suppose we have the function $f(x)=\frac{1}{x}$ defined on $(0,1)$ and we want to prove that $f(z)=2$ for some $z\in (0,1]$ using the Intermediate Value Theorem (of course, it would be easier to say $z=\frac{1}{2}$, but that's not the idea here).

Well, we argue as follows:

Consider the function restricted to $[\frac{1}{4},1]$, since $f(1)=1<2<4=f(1/4)$, by IVT there is some $z\in [\frac{1}{4},1]\subseteq (0,1]$ such that $f(z)=2$.

(Note all of this argument requires continuity of $f$).