Let $(A_n)_{n\geq 0}$ be a collection of rings, with compatible maps $f_{ij}:A_j\to A_i$ for $i\leq j$, and we let $A=\varprojlim\limits_nA_n$, with the canonical projections $\pi_i:A\to A_i$.
If $B\subseteq A$ is a subring of $A$ such that the maps $\pi_i|_{B}:B\to A_i$ are all surjective, must $B=A$? I think the answer is yes, because if there were any element in $A$ that is not in $B$, then it should show up in some $A_i$, but I'm not sure how to make that rigorous.
Let's consider a standard example: $A_n=\Bbb Z/p^n\Bbb Z$ with $p$ prime and the $f_{ij}$ the usual projection maps. The inverse limit is $A=\Bbb Z_p$, the $p$-adic integers.
Let $B=\Bbb Z\subset \Bbb Z_p$. Then $B$ maps onto each $A_n=\Bbb Z/p^n\Bbb Z$.