Let $I$ be any directed poset and $\{X_i,\phi_{ij},I\}$ be an inverse system of profinite spaces with continuous maps $\phi_{ij}\colon X_i \to X_j$ whenever $i\geq j$ in $I$. Let $$X = \varprojlim_{i \in I} X_i$$ be their inverse limit. We may assume that each $X_i$ is a discrete finite space. Let $\phi_i\colon X \to X_i$ be the canonical projection. I'm trying to prove the following claim:
Even if the maps $\phi_i$ are not surjective for general $i$, for every $i \in I$ there exists a $k\in I$ with $k\geq i$ satisfying $\phi_{ki}(X_k) \subseteq \phi_i(X) \subseteq X_i$.
I'm out of clues. Apparently, the hypothesis "$X_i$ is a discrete finite space for all $i$" is essential here, but I'm not really sure how to use it. I've tried to build a contradiction by noting that if $\phi_i$ is not surjective, then $i$ is not a maximal element in $I$ and then finding an element in $X$ whose $i$-th coordinate is not compatible, but this does not seem to be working out.
EDIT: Firstly, I've worked out that we may suppose that $i$ is the lowest element in $I$, for the subset $J = \{j \in I\mid j \geq i\}$ is trivially cofinal in $I$. Working by contradiction, this allows one to construct an inverse system $\{A_j,\phi_{jk},J\}$ with $A_j\subseteq X_j$ for all $j \in J$ satisfying $\phi_{ji}(A_j)\cap \phi_i(X) = \varnothing$. If one manages to prove that $A = \varprojlim_{j \in J} A_j$ is non-empty, it defines an element of $X$ contradicting $A_i \cap \phi_i(X) = \varnothing$. The maps $\phi_{ij}$ need not to be surjective either, so I'm not sure how to go about proving that this limit is non-empty.
On the other hand, I think a bit of context is due. This claim is the last part of the proof of Lemma 2.1.5 in Luis Ribes' Profinite Graphs, to prove that a finite quotient of an inverse limit of profinite graphs factors trough some graph appearing the limit itself. The equivalent lemma for profinite spaces in general is Lemma 1.1.16 in Luis Ribes and Pavel Zalesskii's book Profinite Groups. On the general context of profinite spaces, the claim highlighted is not necessary because the factor map desired doesn't need to preserve any extra structure beyond the topology. The argument in the latter book cannot be directly translated to the graph world, as the factor map constructed there wouldn't preserve the graph structure. Given that the maps $\varphi_{ki}$ are quasimorphisms of graphs, the claim in question would then be sufficient. So while I'm interested in the context of graphs, I believe that the claim is also valid in the general context of profinite spaces (just think that we can assigning to each $X_i$ the trivial graph structure: no edges, only vertices).
EDIT 2 The hypothesis that the maps in the inverse system are surjective for it to be non-empty are not necessary (see, for instance, Proposition 1.1.4 in Profinite Groups). Thus, I believe that the construction $A$ above proves the claim.
This is a compactness argument. There are various ways to phrase it, but here is one. Let's prove it under your reduction that all the $X_i$ are finite discrete spaces.
Claim: Let $a\in X_i$. If for all $k\geq i$, $\phi_{ki}^{-1}(\{a\})\neq \emptyset$, then $\phi_i^{-1}(\{a\})\neq \emptyset$.
Proof: Consider the space $P = \prod_{j\in I} X_j$. As a product of compact spaces, $P$ is compact by Tychonoff's theorem. For all $k\geq i$, let's say a sequence $(x_j)_{j\in I}\in P$ is $k$-good if $\phi_{kj}(x_k) = x_j$ for all $j\leq k$, and $\phi_{ki}(x_k) = x_i = a$. Let $C_k\subseteq P$ be the set of all $k$-good sequences. Then $C_k$ is closed and non-empty by our hypothesis that $\phi_{ki}^{-1}(a)\neq \emptyset$.
For any finitely many $k_1,\dots,k_n\geq i$, let $k\in I$ be some element with $k\geq k_m$ for all $1\leq m\leq n$, and note that $C_k \subseteq C_{k_1}\cap \dots \cap C_{k_n}$. Since $C_k$ is nonempty, the finite intersection of the $C_{k_m}$ is nonempty. By compactness, $\bigcap_{k\geq i} C_k$ is nonempty. Let $(x_j)_{j\in I} \in \bigcap_{k\geq i} C_k$. This sequence defines an element $x\in \varprojlim_{i\in I} X_i$ such that $\phi_i(x) = a$. $\square$
Having proven the claim, let $a_1,\dots,a_n$ enumerate the finitely many elements of $X_i\setminus \phi_i(X)$. Since for each $a_m$, $\phi_i^{-1}(\{a_m\}) = \emptyset$, there is some $k_m\geq i$ such that $\phi^{-1}_{k_mi}(\{a_m\}) = \emptyset$. Let $k\in I$ be some element with $k\geq k_m$ for all $1\leq m \leq n$. Since $\phi_{ki}$ factors as $\phi_{k_mi}\circ \phi_{kk_m}$ for all $m$, we have $a_m\notin \phi_{ki}(X_k)$ for all $m$. Thus $\phi_{ki}(X_k)\subseteq \phi_i(X)$, as desired.
It's also possible to view this through the lens of Stone duality. A codirected limit $X = \varprojlim X_i$ of profinite (Stone) spaces corresponds by Stone duality to a directed colimit $B = \varinjlim B_i$ of Boolean algebras, and a failure of surjectivity in the structure maps $X\to X_i$ corresponds to a failure of injectivity in the structure maps $B_i\to B$. Now it's well-known that in the category of Boolean algebras (and more generally the category of algebras in any equationally axiomatizable class), the map $B_i\to B$ collapses two elements if and only if they are collapsed in one of the maps $B_i\to B_j$ in the diagram. This is the analogue of the claim above.