And $\forall n \in \mathbb{N},$ let $ X_n $ be a non-empty set and $f_n: X_{n+1} \rightarrow X_n.$ And let $\forall n \in \mathbb{N},\operatorname{Pr}_n : \prod_{i\in N}X_i \rightarrow X_n $ be the canonical projection. (i.e. $n$-th component of the given element.)
Let $X=\{x\in \prod_{i\in \mathbb{N}}X_i : \forall n\in \mathbb{N}, \operatorname{Pr}_n(x)=f_{n}(\operatorname{Pr}_{n+1}(x)) \}$.
Then, is it true that ($\forall A\subset X, \forall B\subset X,\space ((\forall a\in \mathbb{N}, \operatorname{Pr}_a(A)=\operatorname{Pr}_a(B) )\rightarrow A=B))$?
For any $x_1, x_2 \in X$, if $\space \exists n\in \mathbb{N}$ s.t. $\operatorname{Pr}_n(x_1)=\operatorname{Pr}_n(x_2),$ then $\forall m \le n, \operatorname{Pr}_m(x_1)=\operatorname{Pr}_m(x_2).$
i.e. all of the formal components of the sequence of $(X_i)_{i\in \mathbb{N}}$ is determined uniquely by the latter component.
For trivial example, let $X_n=\mathbb{N}, f_n : \mathbb{N} \rightarrow \mathbb{N}$ be the injective function like an identity map. Then, any components of $x \in \prod_{i\in \mathbb{N}}$ is deternmined by its 1st component, so it's true for this example.
The motivation of this problem was from the concept of an inverse system, and the above problem is the simpler one of the more general problem and I guess the original problem is false, so I tried to find a counterexample with more simpler problem with which more ones will feel familiar.
For the reference, the more general problem is like this.
Original Problem : Let $I$ be a directed index set, (i.e. $\forall a, b\in I, \exists c \in I, a\le c\wedge b\le c.$)
$\forall i \in I, \space X_{i}$ be a non-empty set.
And $\forall a\le b\in I,$ let $f_{ab} : X_{b} \rightarrow X_{a}$ s.t. $\forall x\le y\le z\in I,\space f_{xy}\circ f_{yz}=f_{xz}$.
And let $\forall a \in I, \operatorname{Pr}_a : \prod_{i\in I}X_i \rightarrow X_a $ be the canonical projection. (i.e. a-th component of the given element and $\forall x, y\in \prod_{i\in I}X_i, (x=y \iff \forall a\in I, \operatorname{Pr}_a(x)=\operatorname{Pr}_a(y))$)
Let $X=\{x\in \prod_{i\in I}X_i : \forall a\le b\in I, \operatorname{Pr}_a(x)=f_{ab}(\operatorname{Pr}_b(x)) \}$.
Then, prove or disprove that $\forall A\subset X, \forall B\subset X,\space (\forall a\in I, \operatorname{Pr}_a(A)=\operatorname{Pr}_a(B)\rightarrow A=B)$
where $\operatorname{Pr}_a(A)=\{ \operatorname{Pr}_a(x):x\in A\}, \operatorname{Pr}_a(B)=\{ \operatorname{Pr}_a(x):x\in B\}.$
If I understand your problem correctly, then no.
Let $X_n$ be the set of all binary sequences of length $n$ (i.e., $X_n$ is the set of all functions from $\{0,1,\cdots, n-1\}$ to $\{0,1\}$.) You can see that $X$ is isomorphic with the set of all infinite binary sequences, and $\operatorname{Pr}_n$ corresponds to the function that cuts off the sequence to that of length $n$. Now consider
(If you are familiar with $2$-adic integers, you may identify $A$ with the set of rational $2$-adic integers, and $B$ with the irrational $2$-adic integers. The idea is that we cannot distinguish the set of rationals and irrationals by just looking at some finite digits of their elements.)
Then $A\neq B$. However, we can see that $\operatorname{Pr}_n A = \operatorname{Pr}_n B$ is the set of all binary sequences of length $n$.