$\langle Tx,x\rangle$ is continuous if and only if that $T$ is continuous. $H$ is a Hilbert space , $T:H \longrightarrow H$ is a linear map , use $T$ to define a functional
$$\phi_T:H \longrightarrow \mathbb C \quad, \quad\quad\phi_T(x)=\langle Tx,x\rangle$$
Then  $\phi_T $ is continuous if and only if $T$ is continuous.
"$\Longleftarrow$"Use the continuity of inner product.
"$\Longrightarrow$"Feng gives a very beautiful proof below,using The Closed Graph Theorem.

Hint: Using the polar identity, we can prove that for any $z\in H$, the map
$$f_z: H\to\mathbb C, \qquad x\mapsto \langle Tx, z\rangle$$
is continuous. Then try to prove that $T$ is a closed operator.
The proof of the closedness of $T$:

 If $x_n\to x $ and $Tx_n\to y$, then for any $z\in H$ we have $\langle Tx_n, z\rangle \to \langle y,z\rangle$ and $\langle Tx_n, z\rangle= f_z(x_n)\to f_z(x)=\langle Tx,z\rangle$. Hence $\langle y,z\rangle=\langle Tx,z\rangle$ for all $z\in H$ and thus $Tx=y$.

