= Solution
Write $\lambda=a+ib$ with $b\ne0$. Since $aI-T$ is self-adjoint, $\langle(aI-T)v,v\rangle$ is real. The cross term in the squared <norm> therefore vanishes:
$$
\|(\lambda I-T)v\|^2=\|(aI-T)v\|^2+b^2\|v\|^2=\|(\overline\lambda I-T)v\|^2.
$$
This proves the equality and lower bound, and makes both $S=\lambda I-T$ and $S^*=\overline\lambda I-T$ <injective>.
If $Sv_n$ converges, the lower bound shows $v_n$ is Cauchy. <Completeness> gives $v_n\to v$ and boundedness gives $Sv_n\to Sv$, so the range is closed. Also $(\operatorname{ran}S)^\perp=\ker S^*=0$, so the range is <dense>. Being both closed and <dense>, it equals $V$. Hence $S$ is bijective, and its lower bound gives
$$
\boxed{\|(\lambda I-T)^{-1}\|\leq|\operatorname{Im}\lambda|^{-1},\qquad\sigma(T)\subseteq\mathbb R.}
$$
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