Remove an eigenvalue from the spectrum of an operator Let $A$ be a bounded linear operator on a $\mathbb R$-Hilbert space, $\sigma(A)$ denote the spectrum of $A$, $\lambda\in\sigma(A)$ be an eigenvalue of $A$ and $$A_\lambda:=\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}.$$

How can we show that $\lambda$ is not an eigenvalue of $A_\lambda$?

First of all, since ${\mathcal N(\lambda-A)}^\perp$ is closed, $A_\lambda:{\mathcal N(\lambda-A)}^\perp\to H$ is again a bounded linear operator. However, is $\mathcal R(A_\lambda)\subseteq{\mathcal N(\lambda-A)}^\perp$ and hence $A_\lambda$ a bounded linear operator on ${\mathcal N(\lambda-A)}^\perp$?
The claim itself should easily follow, since by definition $\mathcal N(\lambda-A_\lambda)=\{0\}$.

Does this even imply that $\lambda\not\in\sigma(A_\lambda)$?

 A: Okay so I am adding a new partial answer as the one I had before was wrong. Turns out that it's not always true that $A_\lambda$ maps $\mathcal{N}(\lambda-A)^{\perp}$ into itself. 
Here's an example. Consider $L$, the left shift on $\ell^{2}(\Bbb{Z}_{>0})$ seen as a real Hilbert space. That is, 
$$
L(x_1, x_2, \ldots):=(x_2, x_3, \ldots)
$$
Then, any $\lambda$ with $|\lambda|<1$ is an eigenvalue for $L$. Fix an eigenvalue $\lambda \in (0,1)$ for $L$. We show that there is a $y \in \mathcal{N}(\lambda-L)^{\perp}$ such that $Ly \not\in \mathcal{N}(\lambda-L)^{\perp}$. We define $y$ as follows
$$
y=(\lambda, -1, 0 , 0 , \ldots)
$$
Now, if $x=(x_1, x_2, \ldots, ) \in \mathcal{N}(\lambda-L)$, that is $Lx=\lambda x$, it follows that $\lambda x_1-x_2=0$. This gives that  $y \in \mathcal{N}(\lambda-L)^{\perp}$. Now, take $z=(\lambda, \lambda^2, \lambda^3, \ldots)$. It's clear that $Lz=\lambda z$ so $z \in \mathcal{N}(\lambda-L)$. However, 
$$
\langle Ly, z \rangle = -\lambda \neq 0
$$
and therefore $Ly \not\in \mathcal{N}(\lambda-L)^{\perp}$.
Remark. Going back to my previous wrong answer, a sufficient condition for $A_\lambda$ to map $\mathcal{N}(\lambda-A)^{\perp}$ into itself is that $\lambda$ is an eigenvalue of $A^*$. This is the case when $A$ is self-adjoint, as you pointed out, but also the case when $A$ is compact. For the example given above, observe that $L^*=R$, where $R$ stands for the right shift given by 
$$
R(x_1, x_2, \ldots)=(0,x_1,x_2, \ldots).
$$
One immediately checks that $R$ has no eigenvalues.
Edit: For your second question, I think that for the example I gave, we also get that $(\lambda-L_\lambda): N(\lambda-L)^{\perp} \to \ell^2(\Bbb{Z}_{>0})$ is not surjective and this will give that $\lambda \in \sigma(L_\lambda)$. For simplicity assume that $\lambda=\frac{1}{2}$. Suppose that there is a $(y_1, y_2, \ldots) \in  N\left( \frac{1}{2}-L\right)^{\perp}$ such that 
$$
\left( \frac{1}{2}-L\right)y=\left(1, \frac{1}{2}, \frac{1}{3}, \ldots \right) \in\ell^2(\Bbb{Z}_{>0})
$$
This gives that $y_n=2\left( \frac{1}{n}+y_{n+1}\right)$ for any $n \in \Bbb{Z}_{>0}$. Hence, 
$$
y_1=\sum_{n=1}^{\infty} \frac{2^n}{n}
$$
But this is not a convergent series. 
