My solution goes as follows. (All norms $||x||$ are norms on $X$)
If $Ax_n-\lambda x_n \rightarrow 0_X$, then $||Ax_n-\lambda x_n||=||(A-\lambda I)x_n||\rightarrow 0$, where $I$ is the identity operator.
This implies that, since $||x_n||=1$, $sup_{||x||=1}||(A-\lambda I)x||=||A-\lambda I||=0$. Is it safe to assume that this implies that $A-\lambda I$ is not invertible? (Like when the determinant of a matrix is $0$ we say it's not invertible)
Edit: Mistake in implying the supremum, it should be infimum, so all of the solution is wrong anyway. But can't seem of thinking on any other way, maybe prove directly that there exists an $x\in X$ such that $Ax=\lambda x$? It should be easy if $(x_n)$ converged (take $x=\lim x_n$).