Representation theorem for quadratic form on Hilbert space

There is indeed a simple proof using the Riesz representation theorem. First note that replacing $x$ by $\lambda^{-1}x$ and $y$ by $\lambda y$ in $\lvert \Psi(x,y)\rvert\leq K(\lVert x\rVert^2+\lVert y\rVert^2)$, you get $\lVert \Psi(x,y)\rvert\leq K(\lambda^{-2}\lVert x\rVert^2+\lambda^2\lVert y\rVert^2)$. With $\lambda=\lVert x\rVert^{1/2}\lVert y\rVert^{-1/2}$ this gives $$ \lvert \Psi(x,y)\rvert\leq 2K\lVert x\rVert \lVert y\rVert. $$ Thus for every $x\in H$ there exists $A(x)\in H$ such that $\Psi(x,y)=\langle A(x),y\rangle$ for $y\in H$ by the Riesz representation theorem. Since $\Psi$ is sesquilinear, the map $x\mapsto A(x)$ is linear, and moreover, $$ \lVert A(x)\rVert=\sup_{\lVert y\rVert=1}\lvert \Psi(x,y)\rvert\leq 2K\lVert x\rVert, $$ so that $A$ is also bounded.

I think this is a fine question for mathoverflow. There does indeed seem to be a convergence issue. However, it can be finessed by restricting to the span of some finite subset of the basis. Then we are working on a finite dimensional space and convergence is trivial. Next, use the uniqueness of $A$ to show that when we pass to a larger finite subset the values $\langle Ax,x\rangle$ do not change. We can also use (1) to get a uniform bound on the norms of the partial versions of $A$, so that they do ultimately yield a bounded operator on all of $H$.