Proof of the spectral theorem for compact operators in Banach spaces Let $X$ be a Banach space over $\mathbb{C}$ and let $T \in \mathcal{K}(X)$ be a compact operator. I want to prove the theorem that every $0 \not= \lambda \in \sigma(T),$ where $\sigma(T)$ denotes the spectrum of $T,$ is an eigenvalue of $T$ using the following results, which I've already proven. I think I'm close but I haven't been able to finish.

1) If $0 \not= \lambda \in \sigma(T),$ then either $\lambda$ is an eigenvalue of $T$ or $\lambda$ is an eigenvalue of its adjoint $T^*.$ Also, if $\lambda$ is not an eigenvalue of $T,$ there exists $\delta > 0$ such that $\Vert Tx - \lambda x \Vert \geq \delta \Vert x \Vert$ for all $x \in X.$
2) The only possible accumulation point of $\sigma(T)$ is 0.
3) If $A$ is a Banach algebra with unit $e,$ $x \in A$ and $\lambda_n$ is a sequence in $\mathbb{C} \setminus \sigma(x)$ with $\lambda_n \to \lambda \in  \sigma(x),$ then $\Vert (x - \lambda_n e)^{-1} \Vert \to +\infty.$

Now I have tried defining $\mu \mapsto \Vert (T - \mu I)^{-1} \Vert$ in a punctured neighbourhood of $\lambda$. My purpose is to derive a contradiction with 3) taking a sequence converging to $\lambda$ and assuming $\lambda$ is not an eigenvalue, but as I said, I'm stuck.
Any help will be appreciated.
 A: $x-\lambda e=x-\lambda_n e +(\lambda_n-\lambda) e$. Hence $(x-\lambda_n e)^{-1} (x-\lambda e)=e+(x-\lambda_n e)^{-1}(\lambda_n-\lambda) e$. If the conclusion fails then the last term tends to $0$ along  a subsequence. But this makes $\|(x-\lambda_n e)^{-1} (x-\lambda e)-e\|<1$ for some $n$. Using the fact  that $y$ is invertible if $\|y-e\|<1$ we see that $(x-\lambda_n e)^{-1} (x-\lambda e)$ is invertible. But then $x-\lambda e$ is invertible. This contradicts the fact that $\lambda \in \sigma (x)$. 
Here is the final argument: suppose $\lambda \neq 0$ is in $\sigma(x)$ and, if possible suppose $\lambda$ is not an eigen value. By 1) we get $\|Tx-\lambda x\|\geq \delta \|x\|$. It is easy to see from this that if $\lambda_n \to \lambda, \lambda \notin \sigma(x)$ then $\|(Tx-\lambda_n x)^{-1}\|\leq \frac  2 {\delta}$ contradicting (c). We have proved that no sequence on the complement of $\sigma (x)$ can converge to $\lambda$. This means $\lambda$ is an interior point of the spectrum. Obviously interior points are limit points of the spectrum. But by (2) $\lambda$ is not a limit point. This finishes the proof.
