Spectrum of a bounded operator and Liouville's theorem

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Let $X$ be a Banach space and $A:X\to X$ be a bounded operator. We know that the spectrum of $A$ is not empty, because otherwise we find a contradiction by using the holomorphy of the function $\lambda \mapsto R(\lambda,A)$ and the Banach version of Liouville's theorem.

I am just asking if there's another approach to prove that the spectrum of $A$ is not empty without using Liouville's theorem.

This will make me better understand the power and beauty inside Liouville's theorem and complex analysis in general.