I have an hint in order to prove the statement in the title:
If $\kappa$ is an $\eta$-extendible cardinal then it is $\delta$-extendible for every $\delta <\eta$.
The hint is to prove that the elementary embedding $j:V_{\kappa+\eta}\longrightarrow V_\beta$ (which exists for some $\beta$ by hypothesis) restricts to an elementary embedding $V_{\kappa+\delta}\longrightarrow V_\beta$.
From this I can conclude, but why is the hint true?
Thank you in advance for any help.
Let $$ j \colon V_{\kappa + \eta} \to V_{\beta} $$ be an elementary embedding and let $\beta ' := j(\kappa + \delta) < \beta$. I claim that $$ j \restriction V_{\kappa + \delta} \colon V_{\kappa + \delta} \to V_{\beta'} $$ is an elementary embedding. Why? Let $\vec{x} \in V_{\kappa + \delta}$ and let $\phi$ be a formula of set theory. Then $$ \begin{align*} V_{\kappa + \delta} \models \phi[\vec{x}] & \iff V_{\kappa + \eta} \models "V_{\kappa + \delta} \models \phi[\vec{x}] " \\ & \iff V_{\beta} \models " j(V_{\kappa + \delta}) \models \phi[j(\vec{x})]" \\ & \iff V_{\beta'} \models \phi[j(\vec{x})]. \end{align*} $$ The last equivalence follows because $j(V_{\kappa + \delta}) = V_{\beta'}$ using both elementarity and the fact that $V_{\beta}$ computes $V_{\beta'}$ correctly. The last part needs a bit justification, if you don't already know this fact.