The question is as follows:
Let $A \in L(X)$, $X$ normed space, $\lambda \in \mathbb{C}.$ Assume that there exists a sequence $\{ x_n \}_{n \in \mathbb{N}}$ in $X$ such that $\| x_n \| = 1$ for all $n$, and $Ax_n - \lambda x_n \to 0$ as $n \to + \infty.$ Prove that $\lambda \in \sigma(A) = \{ \lambda \in \mathbb{C} \mid A - \lambda I ~ \text{is not invertible} \}.$
$\textbf{An idea:}$
I want to show that $\| A - \lambda I \| $ goes to zero and then it will not be invertible.
WE have \begin{align} \| A - \lambda I \| &= \| A - A x_n + A x_n - \lambda x_n + \lambda x_n - \lambda I \| \\& \le \| A - A x_n \| + \| A x_n - \lambda x_n \| + \| \lambda x_n - \lambda I \| \\& \le \|A\| \| I - x_n \| + \| A x_n - \lambda x_n \| + |\lambda | \| I - x_n \| \\& = 0 \end{align}
Then we have that $A - \lambda I$ is not invertible.
Please let me know if I am wrong?
Thanks!
Your proof is wrong. It doesn't make sense to talk about $||A-Ax_n||$.
The problem you gave is a general Banach space problem: let $T:X \to X$ be a bounded linear map such that there exist $(x_n)_n \in X, ||x_n|| = 1$ with $Tx_n \to 0$. Then $T$ is not invertible. [Note in your problem, $T= A-I$].
The proof of this general problem is as follows. If $T$ were invertible in $L(X)$, then by definition it is bounded. So, there is some $C \ge 0$ so that $||T^{-1}x|| \le C ||x||$ for all $x \in X$. Since $T$ is invertible, it is surjective, so we may replace $x$ with $Tx$ to get $||x|| \le C ||Tx||$ for all $x \in X$. Using this for $x=x_n$ gives $1 \le C ||Tx_n||$. But this is a contradiction for large $n.