I started out by rearranging the equation: $$e^x-y^2 cos x = 0$$ Set $f(x) = e^x-y^2 cos x$, and $k = 0$. By the intermediate value theorem, if $f(x)$ is continuous on $[0,\pi/2)$, and $f(0) \leq f(\pi/2)$, then $f(0)\leq 0 \leq f(\pi/2) \Rightarrow 0 = f(c)$ for some $c \in [0, \pi/2)$; in other words, the equation has a solution for $x$ lying in $[0, \pi/2)$.
It is easy to observe that $f(x)$ is continuous on $[0, \pi/2)$. We know that $$f(0) = 1-y^2$$ and $$f(\pi/2) = e^{\pi/2}$$ So for values of $y \geq 1$, $$f(0)\leq 0 \leq f(\pi/2)$$ which implies $$0 = f(c)$$ For some $c \in [0, \pi/2)$, and the proof is complete.
Is this proof legitimate? Also, why the open interval? I think it makes no difference in terms of the reasoning.
Thank you!
Your proof is mostly correct. To be more clear, you want to separate the $y$ into two cases: $y = 1$, then $f(0) = 0$, thus $x=0$ is indeed a solution. If $y > 1$, then $f(0) = 1 -y^2 < 0, f(\pi/2) = e^{\pi/2} > 0$, then IVT theorem says $f(c) = 0$ for some $c \in (0,\pi/2)$ and $c$ is a solution.