Let $\lambda > \aleph_{0}$ and M a $\lambda$-saturated structure. Show that for every non-algebraic $\phi(x,a)$ with parameters from $M$ there exists a sequence of indiscernibles $I \subset \phi(M,a)$ of cardinality $\lambda$.
As $\phi(x,a)$ is not as algebraic then $\phi(x,a)$ is infinite, because it must be sized $\lambda$?
Whenever any $\lambda$-saturated model $M$ I can find a succession of indiscernible of size $\lambda$?
Is it possible to remove the saturation assumption from the previous statement?
It is not because it must be sized $\lambda$. It is infinite by the definition of a non-algebraic formula. And it is not $\phi(x,a)$ that is infinite, but $\phi(M,a)$.
You can construct recursively a sequence of indiscernibles of size $\lambda$. First, you can find one of length $\omega$ by Ramsey and compactness and $\aleph_1$-saturation. Then, you can add one more by compactness and $\lambda$-saturation, until you have a sequence of length $\lambda$.
No, it is not possible to remove the saturation from the previous statement. For example, if $M$ is countable, there are no uncountable (nonconstant) indiscernible sequences in it.