TITLE: Eigenvalues of compact operator don't have nonzero accumulation points
QUESTION [6 upvotes]: In the book 

Elements of the theory of functions and functional analysis 

of Kolmogorov and Fomin, there is a proof of the following theorem,
Every compact operator $A$ on a Banach space $E$ has for arbitrary $\rho>0$ only a finite number of linearly independent eigenvectors which correspond to the eigenvalues whose absolute value are greater than $\rho$.
The proof that it gives is later in the errata part of the book as wrong, but it doesn't say why and I don't know what is wrong with the following similar argument,
Assume that the theorem is false, that is, there exist a $\rho$ and a sequence $\{x_n\}$ of linearly independent vectors with $A x_n = \lambda_n x_n$ and $|\lambda_n|>\rho$. We can restrict $A$ to the span of this sequence, it is easy to check that it has a bounded inverse (if $x=a_1x_1+ \cdots +a_k x_k$ then $A^{-1}x=a_1 x_1/\lambda_1 + \cdots + a_kx_k/\lambda_k$ and $||A^{-1}|| \leq 1/\rho$) therefore on this space we have a compact operator with bounded inverse and consequently the identity map in this space is compact , arriving to a contradiction since the space is infinite dimensional.

REPLY [2 votes]: @inquisitor: I looked at a translation of Kolmogorov-Fomin into Spanish, can't differ too much from the original. Here is how they prove it, with some smoothing.
You take $E$ normed space and $A$ a linear operator, $x_n\in E$  linearly independent  so that $A x_n = \lambda_n \cdot x_n$,  and $|\lambda_n| \ge \delta > 0$. We will show that $A$ is not a compact operator. 
First, some standard facts:

If $F$ is a closed subspace of $E$ and $x \not \in F$ then there exist $\alpha \in \mathbb{K}$ and $y \in F$ so that $||\alpha x + y || = 1$ and $\text{d}( \alpha x + y, F) \ge \frac{1}{2}$
If $F$ is a subspace of $E$ and $A x = \lambda x$ then for all vectors $z$ in the subspace $\mathbb{K} \cdot x + F$ we have $A z = \lambda z + g$ for some $g \in F$. 

Consider now the sequence of finite dimensional subspaces $F_n = (x_1, \ldots, x_n)$. Take an $n$ and apply $1.$ to $x_n$ and $F_{n-1}$. We get $y_n\colon = \alpha_n x_n + f_{n-1}$ so that 
$||y_n||=1$ and $\text{d}(y_n, F_{n-1}) \ge 1/2$.
Note that the span of $y_1, \ldots , y_n$ is still $F_n$, for all $n$ (by induction). Moreover, from $2.$ we have $A y_n = \lambda_n y_n + g_n $ for some $g_n \in F_{n-1}$.
We are done now. For $n>m$ we have 
$(Ay_n - A y_m) = \lambda_n\cdot (y_n - h)$ where $h \in F_{n-1}$ and so 
$||Ay_n - A y_m|| = |\lambda_n| \cdot ||y_n - h||\ge \frac{1}{2} |\lambda_n|\ge \frac{1}{2} \delta$ 
while $||y_n|| =1$ for all $n$.