I am trying to prove that a non zero limit ordinal $\gamma$ always has a preceding limit ordinal (counting zero as a limit ordinal).

I am not sure if this is true but if it is $\bigcup\{\delta\mid\delta < \gamma \text{ and } \delta \text{ is a limit ordinal} \} $ would be the candidate.

I can prove a union of limit ordinals is a limit ordinal. It's proving the above limit ordinal is less than $\gamma$ that is unclear to me.