The Intermediate Value Theorem states:
Theorem. Let $f(x)$ be a function that is continuous on $[a,b]$. If $k$ is a number between $f(a)$ and $f(b)$, then there exists (at least) one $c$, $a\lt c \lt b$, such that $f(c) = k$.
I often tell my Calculus students to think of the Intermediate Value Theorem as the "Goldilocks Theorem". You are trying to find a point where $f$ takes the value $k$. If one of $f(a)$ and $f(b)$ is "too much", and the other is "too little", then somewhere in between is a point where $f$ is "just right."
The most common (but by no means only) use of the Intermediate Value Theorem is to find points where $f(x)$ is equal to $0$. If you can find a point where $f(x)$ is positive and one where $f(x)$ is negative, then somewhere in between there will be a point where $f(x)$ is exactly $0$ (provided $f$ is continuous).
In the case at hand, you want to find a solution to $e^{-x^2}=x$. So we can take $f(x) = e^{-x^2}-x$; we are looking for a point where $f(x)=0$. This function is continuous everywhere. Since $f(0) = e^{0}-0 = 1$ and $f(1) = e^{-1}-1 \lt 0$, we have:
$f(x)$ is continuous on $[0,1]$, it is positive at $0$ (value is "too much"), and it is negative at $1$ (value is "too little"). Therefore, by the Intermediate Value Theorem, there is at least one point $c$ somewhere in $(0,1)$ where $f(c)=0$ ("just right").
But if $f(c)=0$, then that means that $0 = e^{-c^2}-c$, so $e^{-c^2}=c$. That is, the equation $e^{-x^2} = x$ has at least one solution in $(0,1)$.