In a Hilbert space $X$, $T \in B(X)$, $|\lambda|=\lVert T \rVert$. Prove that $Im(\lambda I - T)+Ker(\lambda I-T)$ is dense in $X$

Suppose that $x$ is orthogonal to this space, and let's also normalize, $\|x\|=1$. Then, in particular, $x\in R(T-\lambda)^{\perp}=N(T^*-\overline{\lambda})$, so $$ \overline{\lambda} \langle x, Tx \rangle = \langle x, TT^* x\rangle = \|T^*x\|^2 = |\lambda|^2 $$ and thus $\langle x, Tx \rangle =\lambda$. On the other hand, $$ |\langle x, Tx \rangle | \le \|Tx\| \le |\lambda| \quad( = |\langle x, Tx \rangle |) . $$ Equality in the Cauchy-Schwarz inequality means that the vectors are linearly dependent, so $Tx = cx$, and only $c=\lambda$ works here.

So, in conclusion $x\in N(T-\lambda)$, but we also assumed that $x$ is orthogonal to this space, so $x=0$. (This whole argument assumes that $\lambda\not= 0$; of course, the claim is trivial if $T=0$.)