Show that $\psi:L(E,L(E,F))\rightarrow L^2(E,F)$ given by $[\psi(T)](u,v)=[T(u)](v)$ is a linear homeomorphism

We follow the steps asked. Beginning with a proof that $L^2(E,F)$ is complete.

First let $T_n$ be cauchy in $L^2(E,F)$. It follows for any $u,v\in E$ that
$$\|T_n(u,v)-T_m(u,v)\|=\|(T_n-T_m)(u,v)\| ≤ \|T_n -T_m\|\, \|u\|\,\|v\|$$
and so $T_n(u,v)$ is cauchy in $F$, thus it admits a limit. We define a map $T:E\times E\to F$ by:
$$T(u,v):=\lim_n T_n(u,v)$$
for $u,v\in E$ arbitrary. It is trivial to check this is bilinear:
$$T(\lambda u_1 + u_2 , v) = \lim_n T_n(\lambda u_1+u_2,v) = \lim_n (\lambda T_n(u_1,v)+T_n(u_2,v))=\lambda \lim_n T_n(u_1,v) +\lim_n T_n(u_2,v) = \lambda T(u_1,v)+T(u_2,v)$$
giving linearity in the first component, the proof for the second component is the same.

Next we check that $T$ is also continuous, for this we note that since $T_n$ is cauchy there is some $L>0$ so that $\|T_n\|<L$ for all $n$. Then:
$$\|T(u,v)\| = \|\lim_n T_n(u,v)\| = \lim_n \|T_n(u,v)\|≤ \lim_n (L\|u\|\,\|v\|)=L\|u\|\,\|v\|$$
which implies continuity of $T$, since if $(u_n, v_n)\to (u,v)$ you get
$$\|T(u_n,v_n)-T(u,v)\| ≤ \|T(u_n-u, v_n)\|+\|T(u, v_n-v)\|≤ L (\|u_n -u\|\,\|v_n\|+\|u\|\,\|v_n-v\|)$$
where the right-hand side converges to $0$.

So we have proven that any cuachy sequence in $L^2(E,F)$ admits a limit in $L^2(E,F)$. So $L^2(E,F)$ is complete - ie Banach.

Next we check that the map $\Psi : L(E, L(E,F))\to L^2(E,F)$ is surjective. So suppose $T\in L^2(E,F)$. For each $u\in E$ you have that the map
$$A_u: E\to F, \qquad v\mapsto A_u(v) := T(u,v)$$
is continuous + linear map $v$. Hence we get a map
$$A: E\to L(E,F),\qquad u\mapsto A_u$$
we check now that this map is continuous and linear in $u$.

Linearity is obvious, since $A_{\lambda u_1 + u_2}(v) = T(\lambda u_1 + u_2, v) = \lambda T(u_1,v)+T(u_2, v) = (\lambda A_{u_1}+ A_{u_2})(v)$ holds for all $v$ and hence $A_{\lambda u_1 + u_2} = \lambda A_{u_1}+A_{u_2}$. For continuity if $u_n\to u$ you have:
$$\|A_{u_n}-A_u\| = \sup_{\|v\|≤1}\|(A_{u_n}-A_u)(v)\| = \sup_{\|v\|≤1}\|T(u_n,v)-T(u,v)\| ≤\|T\|\,\|u_n-u\|$$
which converges to $0$. Hence you also get continuity and $A\in L(E,L(E,F))$.

Finally we check that $\Psi(A)=T$, this is immediate:
$$\Psi(A)(u,v)= [A_u](v)=T(u,v)$$

So the map $\Psi$ is continous, linear + surjective between Banach spaces. We finally check injectivity to get that its a homeomorphism by the open mapping theorem.

Suppose $\Psi(A)=0$, then for all $u\in E$ you have that the map
$$v\mapsto [A(u)](v)=0$$
hence $A(u)$ is the zero map for all $u$, ie $A=0$. So $\ker(\Psi)=\{0\}$ and $\Psi$ is injective.