= Solution
Again put $A=\lambda I-T$. Surjectivity gives, by <image-kernel orthogonality for an adjoint>,
$$
\ker A^*=(\operatorname{Im}A)^\perp=X^\perp=\{0\}.
$$
If $\lambda$ is real, $A=A^*$, so $\ker A=\{0\}$. If $\lambda$ is nonreal and $Ax=0$, then $Tx=\lambda x$; reality of $\langle Tx,x\rangle=\lambda\|x\|^2$ again implies $x=0$. Thus $A$ is injective as well as surjective, without needing to assume separately that $\lambda$ is not an <eigenvalue>.
Since $A$ is a <bounded linear operator> on the <Banach space> $X$, the <bounded inverse theorem> makes $A^{-1}$ bounded. Therefore
$$
\boxed{\lambda\in\rho(T),\qquad(\lambda I-T)^{-1}\in\mathcal B(X).}
$$
This is exactly membership in the <resolvent set of an operator>.
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