The following is exercise 7.3.1. of Tent and Ziegler's textbook:
Let $T$ be simple, with a monster model $\mathfrak{U}$, and let $\pi(v,w)$ be a collection of $\mathcal{L}_A$-formulas with $b\in\mathfrak{U}$ such that $\pi(v,b)$ is consistent and does not fork over $A$. If $\mathcal{I}$ is an infinite sequence of $A$-indiscernibles containing $b$, then there is a realization $c$ of $\pi(v,b)$ such that $c\perp_A\mathcal{I}$ and such that $\mathcal{I}$ is indiscernible over $A\cup\{c\}$.
Here is my solution: By compactness, it suffices to assume that $\mathcal{I}$ has order type $\omega$, so let $\mathcal{I}=(b_i)_{i\in\omega}$, with $b_0=b$. Now, let $p(v)\in S_n(A\cup\mathcal{I})$ be a non-forking extension of $\pi(v,b)$, with a realization $c$. By construction, we have $\mathfrak{U}\models\pi(c,b)$ and $c\perp_A\mathcal{I}$. We claim that $\mathcal{I}$ is $A\cup\{c\}$-indiscernible. To see this, we want to show $$\varphi(v,b_{i_1},\dots,b_{i_k})\leftrightarrow\varphi(v,b_{j_1},\dots,b_{j_k})\in\text{tp}(c\big/A\cup\mathcal{I})$$ for every $i_1<\dots<i_k\in\omega$ and $j_1<\dots<j_k\in\omega$ and every $\mathcal{L}_A$-formula $\varphi$. To show this, let $n$ be maximal among the $i_l$ and $j_l$. Since $c\perp_A\mathcal{I}$, in particular we have $c\perp_A\{b_0,\dots,b_n\}$. Then $\mathcal{I}$ may be considered as an $A$-indiscernible sequence with first element $(b_0,\dots,b_n)$ in a natural way, and so by corollary 7.1.5. we may find some $c_n\in\mathfrak{U}$ with $c_n\equiv_{A\cup\{b_0,\dots,b_n\}}c$ and such that $\mathcal{I}$ is $A\cup\{c_n\}$-indiscernible. In particular, the desired formula lies in $$\text{tp}(c_n\big/A\cup\{b_0,\dots,b_n\})=\text{tp}(c\big/A\cup\{b_0,\dots,b_n\})\subseteq\text{tp}(c\big/A\cup\mathcal{I}),$$ and so we are done.
I can't see any error in what I've written, but I looked in the back of Tent and Ziegler's book, and they give a hint for a (seemingly) more elaborate solution, which first proves that $\bigcup_{i\in\omega}\pi(v,b_i)$ does not fork over $A$ and then takes a non-forking extension of that collection. Perhaps the ideas are not so different, since the proof of lemma 7.1.5. uses a similar union of pushforward types ranging over an indiscernible sequence, but I'm not sure whether there's a reason Tent and Ziegler didn't just use lemma 7.1.5. in their solution to this exercise. In particular, is there a mistake anywhere in my proof?
In your proof, you replace the $A$-indiscernible sequence $\mathcal{I}$ with a "thickening": an $A$-indiscernible sequence $\mathcal{I}^n = (b_{(n+1)k},b_{(n+1)k+1},\dots,b_{(n+1)k+n})_{k\in \omega}$. Then its true that you can find $c_n \equiv_{Ab_0,\dots,b_n} c$ such that $\mathcal{I}^n$ is $Ac_n$-indiscernible. But you cannot conclude from this that $\mathcal{I}$ is $Ac_n$-indiscernible.
For example, $b_0$ and $b_1$ may have different types over $Ac_n$. Your construction only ensures that $b_0$, $b_{n+1}$, $b_{2(n+1)}$, etc. have the same type over $Ac_n$.
In general, it does not follow from $\mathfrak{U}\models\pi(c,b)$ and $c\perp_A\mathcal{I}$ that $\mathcal{I}$ is $A\cup\{c\}$-indiscernible. For example, in the theory of the random graph, suppose $\pi(x,y)$ says $xRy$. We can take $c$ to be any element not in $\mathcal{I}$, with $cRb_0$, and we will have $c\perp_A\mathcal{I}$. In particular, $c$ can be connected by an edge or not to the other elements of $\mathcal{I}$. So you have to do something more elaborate to prove this (in particular, you have to use simplicity somewhere, which your solution does not - the statement is not true for arbitrary theories).