$\lambda$ is in the point spectrum.
By definition it exists a $x\in H$ such that $\Vert x \Vert = 1$ and $Tx = \lambda x$. We have $ (Tx,x) = (\lambda x, x) = \lambda $ so $\lambda$ belongs to $W(T)$.
$\lambda$ is in the residual spectrum.
Then it exists $x\in H$ such that $\Vert x \Vert = 1$ and $x$ is orthogonal to the range of $T - \lambda$.
For each $y\in H$ we have $((T - \lambda)y, x) = 0$, in particular $ 0 = ((T - \lambda)x, x) = (Tx, x) - \lambda $ Also in this case $\lambda$ belongs to $W(T)$.
$\lambda$ is in the continuous spectrum.
$(T- \lambda)^{-1}$ is not bounded, therefore it exists a sequence $\{x_n\}_{n\in \mathbb N}$ of elements of $H$, with $\Vert x_n\Vert = 1$ and $\Vert (T - \lambda)^{-1} x_n\Vert \to \infty$.
Let's consider the sequence $ y_n := \frac {(T - \lambda)^{-1} x_n} {\Vert (T - \lambda)^{-1} x_n \Vert} $ we have $\Vert y_n \Vert = 1$ and $ (T - \lambda)y_n = \frac {x_n} {\Vert (T - \lambda)^{-1} x_n \Vert} \to 0 $ If $\lambda \notin \overline{W(T)}$ then it exists a $M > 0$ such that $\vert \lambda - (Tx, x)\vert > M$ for each $x\in H$, $\Vert x \Vert = 1$. As conseguence $ \Vert (T-\lambda)x \Vert = \Vert (T-\lambda)x \Vert \Vert x \Vert \geq \vert ((T - \lambda)x, x)\vert = \vert \lambda - (Tx, x)\vert > M $ for each $x\in H$, $\Vert x \Vert = 1$. But that contradicts the existence of the sequence $y_n$.