Hellinger-Toeplitz Theorem and Uniform Boundedness Principle [closed]

The question is prove the Hellinger-Toeplitz theorem, using the uniform boundedness principle. All hints are welcome

Let $H$ be a Hilbert space and $T:H\mapsto H$ be a symmetric operator defined everywhere. Consider the family of linear functionals $\lbrace f_x\rbrace_{x\in\mathbb{B}_H}$ where $f_x:\mathbb{B}_H \mapsto\mathbb{K}$ is given by $$f_x(y)=\langle Tx,y\rangle$$ and $\mathbb{B}_H$ denotes the unit ball of $H$.

Take $y\in \mathbb{B}_H$; then $$\sup_{x\in \mathbb{B}_H} |\langle Tx,y\rangle|=\sup_{x\in \mathbb{B}_H} |\langle x,Ty\rangle|<\infty$$ as the inner product defines a linear functional. As $y\in \mathbb{B}_H$ was arbitrary, the Uniform Boundedness Principle applies, so that $$\sup_{x\in \mathbb{B}_H} \sup_{y\in \mathbb{B}_H} |\langle Tx,y\rangle|<\infty$$

Finally note that $$||T||^2=\sup_{x\in \mathbb{B}_H} ||Tx||^2=\sup_{x\in \mathbb{B}_H} |\langle Tx,Tx\rangle|\leq \sup_{x\in \mathbb{B}_H} \sup_{y\in \mathbb{B}_H} |\langle Tx,y\rangle|<\infty$$

Proving $T$ is bounded.