We need
A lemma by Riesz: If $X$ is Banach and $M\subset X$ is a closed subspace then for every $\epsilon>0$ there is $x\in X$ such that $||x||=1$ and $d(x,M)>1-\epsilon$.
Proof: Let $1>\epsilon>0$. Pick $y\in X\setminus M$. Then $d(y,M)=a>0$. There exist $m\in M$ such that $$||y-m||\leq a\left(1+\frac{\epsilon}{1-\epsilon}\right)$$ Put $x:=\frac{y-m}{||y-m||}$. We can check that for $z\in M$ we have $$\begin{align}||x-z||&=\left|\left|\frac{y-m}{||y-m||}-z\right|\right|\\&=||y-m||^{-1}\cdot\left|\left|y-\color{red}{m-||y-m||\cdot z}\right|\right|\\&=||y-m||^{-1}\cdot||y-\color{red}{h}||\\&\geq||y-m||^{-1}\cdot a\ \ \ \ \text{ because }\color{red}{h}\in M\\&\geq 1-\epsilon\end{align}$$
Proving that eigenvalues of compact operators can only accumulate at $0$.
Assume that $\lambda\neq0$ is not an eigenvalue. Then $T:=A-\lambda$ is one-to-one.
Final goal: To show that $T^{-1}:X\to X$ exists and it is bounded. In particular this shows that $\lambda$ can't be a limit of eigenvalues.
Assume the image of $T$ is closed.
Then by the open mapping theorem $T^{-1}:\text{Im}(T)\to X$ exists and its continuous.
If $X_1:=\text{Im}(T)\subsetneq X$ then all spaces $X_{n+1}:=\text{Im}(T)X_{n}$ are different, with each containing the next one.
Now, pick $x_n\in X_n$ such that $||x_n||$, and $d(x_n, X_{n+1})>\frac{1}{2}$. Then, for $m>n$
$$\begin{align}\left|\left|Ax_n-Ax_m\right|\right|&=|\lambda|\cdot\left|\left|x_n+\color{red}{-x_m-\frac{Tx_n-Tx_m}{\lambda}}\right|\right|\\&=|\lambda|\cdot||x_n-\color{red}{x}||\ \ \ \ \text{ where }\color{red}{x}\in X_{n+1}\\&\geq\frac{|\lambda|}{2}\end{align}$$
But this means that the sequence $Ax_n$ can't have convergent subsequences contradicting that $A$ is compact.
Therefore $\text{Im}(T)=X$ and $T^{-1}:X\to X$.
We can finish now because
The image of $T$ is closed.
Assume that $y_n=Tx_n$ converges to $y$.
Case 1: If $x_n$ is bounded then, since $A$ is compact there is a subsequence $x_{n_k}$ such that $Ax_{n_k}$ converges. But $x_{n_k}=\lambda^{-1}(Ax_{n_k}-y_{n_k})$, so $x_{n_k}$ is itself convergent (to some $x$). Then $y=Tx\in\text{Im}(T)$.
Case 2: If $x_n$ is unbounded (we can assume $||x_n||\to\infty$). Then $z_n:=x_n/||x_n||$ has norm $1$ and $Tz_n=y_n/||x_n||\to0$ (because $y_n$ is convergent and therefore bounded). Because $z_n$ is bounded and $A$ is compact, there is a subsequence $z_{n_k}$ such that $Az_{n_k}$ converges. But then $z_{n_k}=\lambda^{-1}(Az_{n_k}-Tz_{n_k})$ is itself convergent (to some $z$). Taking limit in this last equation we see that $\lambda z=Az$. Therefore $z=0$, since we assumed that $\lambda$ is not an eigenvalue. But this can't be because $||z_n||=1$ doesn't allow $z_n\to0$.