Showing that the Spectrum of a Hermitian/Self Adjoint Operator is Real I have this question:

Let $A$ be a Hermitian (self-adjoint) operator on a Hilbert space $H$. Let $\lambda \in \mathbb{C}$ such that
$\operatorname{Im}(\lambda) \neq 0.$ Let  $B = \lambda I - A.$ Prove that
(a) $B$ is 1 - 1.
(b) $B(H)^\perp = \{0\}$.
(c) If $\lim\limits_{n\to\infty}By_n = x$, then $\|x_n - x_m\|^2 \geq (\operatorname{Im}(\lambda))^2\|y_n - y_m\|^2$,
where $x_n = By_n$ for each $n.$
Hence, prove that $\lambda$ is not in the spectrum $\sigma(A).$

I was able to prove part a and b by using definitions and Cauchy-Schwartz inequality. But I could not show part c and  $\lambda \notin \sigma(A)$. Can someone help me for this question?

Note that (b) implies that the range of $B$ is dense. If we can show that it's all of $H$, then we'll know that $B$ is bijective, which means that $\lambda\notin\sigma(A).$ This will follow from (c). Indeed, take $x\in H.$ By density, there exists a sequence $(y_n)$ in $H$ so $By_n\rightarrow x.$ Using part (c), we get that $(y_n)$ is Cauchy, so it converges, say to $y\in H$. By continuity, we get that $By=x$, so $x\in B(H).$ The other inclusion is obvious.
All that's left is to show part (c). Simply observe that 
\begin{align*}
\|x_n-x_m\|^2&=\|By_n-By_m\|^2=\|(\lambda-A)(y_n-y_m)\|^2\\
&=\|(\text{Re}\ \lambda-A)(y_n-y_m)+i\text{Im}\ \lambda(y_n-y_m)\|^2\\
&= \|(\text{Re}\ \lambda-A)(y_n-y_m)\|^2+\|\text{Im}\ \lambda(y_n-y_m)\|^2\\
&\geq |\text{Im}\ \lambda|^2\|y_n-y_m\|^2 
\end{align*} (Self-adjointness was used when moving from the second to the third line.)
