Let $\mathbb{I}=\langle a_{i} \mid i< \omega + \omega\rangle$ be a sequence of indiscernibles over a set $A$. Then $\langle a_{i} \mid \omega \le i< \omega + \omega\rangle$ is an indiscernible sequence over $A \cup \langle a_{i} \mid i< \omega \rangle$.
This is a specific case of a stronger result: Let $(I, <)$ be an infinite linearly ordered set and suppose that exists $p \in I$ such that $\{q \in I \mid p<q\}$ is infinite, if $\langle a_{i} \mid i \in I \rangle$ is a sequence of indiscernible over $A$ then $\langle a_{i} \mid i \in I, p<i\rangle$ is a sequence of indiscernibles over $A \cup \{a_{i} \mid i \in I, i<p \}$.
I do not understand why the sets $\{q \in I \mid p<q\}$ and $\{q \in I \mid p>q\}$ must be both infinite. At first $I$ thought of using a permutation of the set I to order ascending elements of $\langle a_{i} \mid i \in I, p<i\rangle$ and use the hypothesis that $\langle a_{i} \mid i \in I \rangle$ is a sequence of indiscernible over $A$. But apparently is not the right way.
You don't need either set to be infinite. The full result should be
Proof sketch. We need to show that the truth of a formula about $\{b_i\}_{i\in I_0}$ with parameters from $\{b_i\}_{i\in I_1}$ doesn't change when we move the $\{b_i\}_{i\in I_0}$ pieces (preserving order). (The other part is the same argument with the roles of $I_0$ and $I_1$ switched.) This result follows immediately by the indiscernibility of the full sequence, together with the fact that $I_0 < I_1$ means we can't have moved any of the terms "past" the parameters (in $I$'s order).
(Exercise: find an example where the converse fails.)