When is the projective limit non-empty? Let $\{S_i\}_{i \in I}$ be a directed projective system and $S = \varprojlim S_i$. 
1.How to prove that $S$ is non-empty in the following case: all $S_i$ are nonempty compact Hausdorff spaces.
2.I know that when $S_i$ are non-empty sets and $f_{ij}$ are surjective, $S$ may still be empty. But if we furthermore assume that $I$ is countable, then $S$ can be shown to be non-empty. What more can we say about the non-emptyness of $S$?
 A: This answer is to supplement Fred Rohrer's answer, copying out some details extracted from Bourbaki (in writing for instance TG.I.9.6, he was referring to a French edition of Topologie Générale, but I'll just refer to the English language edition of Bourbaki's Set Theory Treatise, starting here, page 205/418). 
Suppose given a directed set $D$ (i.e., a partially ordered set such that any two elements have an upper bound), regarded as a category, and suppose given a functor $F: D^{op} \to Set$ whose value at $\alpha \in D$ is denoted $F_\alpha$, with transition maps denoted $f_{\alpha \beta}: F_\beta \to F_\alpha$ when $\alpha < \beta$. Suppose further that for each $\alpha \in D$, one is given a collection of subsets $S_\alpha \subseteq P(F_\alpha)$, each closed under arbitrary intersections [so vacuously $F_\alpha \in S_\alpha$!] and satisfying the finite intersection property. (For example, if the $F_\alpha$ carry compact Hausdorff space structures, we could take $S_\alpha$ to be the collection of closed subsets of $F_\alpha$.) The finite intersection property could be replaced by: 

If $K \subseteq S_\alpha$ is codirected, i.e., has the property that for all $A, B \in K$ there is $C \in K$ with $C \subseteq A$ and $C \subseteq B$, and if also $\emptyset \notin K$, then $\bigcap K \neq \emptyset$. (**) 

Theorem: Assume that $f_{\alpha\beta}(C) \in S_\alpha$ whenever $C \in S_\beta$, and that $f^{-1}(x) \in S_\beta$ whenever $x \in F_\alpha$. Then for the projective limit $L$ with universal cone $f_\alpha: L \to F_\alpha$, we have $f_\alpha(L) = \bigcap_{\alpha \leq \beta} f_{\alpha\beta}(F_\beta)$. 
Note we have the following corollary. 
Corollary: Under the assumptions of the theorem, if all the $F_\alpha$ are nonempty, then so is $L$. 
This follows from the theorem by choosing any $\alpha \in D$ and taking $K = \{f_{\alpha\beta}(F_\beta): \alpha \leq \beta\}$, noting that $K$ is codirected because $D$ is directed, and applying the highlighted condition (**). (In the exceptional case where $D$ is empty, the projective limit $L$ is the terminal set, hence nonempty.) 
Here is an outline of how Bourbaki proves the theorem. First he sets up a Zorn's lemma situation: let $\Sigma$ be the poset whose elements are tuples of the form $(A_\alpha)_{\alpha \in D}$ such that $A_\alpha \in S_\alpha$ and $A_\alpha \neq \emptyset$ and $f_{\alpha\beta}(A_\beta) \subseteq A_\alpha$ whenever $\alpha \leq \beta$, and with $\Sigma$ partially ordered by  $(A_\alpha) \preceq (B_\alpha)$ if $B_\alpha \subseteq A_\alpha$ for all $\alpha \in D$. (Note the change of direction.) Using the highlighted form of the finite intersection property, he notes that every chain in $(\Sigma, \preceq)$ has an upper bound. Thus $\Sigma$ has a maximal element, by Zorn's lemma. 
He next shows that if $(A_\alpha)$ is maximal, then $f_{\alpha\beta}(A_\beta) = A_\alpha$. In effect he considers $A_{\alpha}' := \bigcap_{\alpha \leq \beta} f_{\alpha\beta}(A_\beta)$ and checks that $(A_{\alpha}')$ satisfies the requirements for belonging to $\Sigma$ and that $f_{\alpha\beta}(A_{\beta}') = A_{\alpha}'$. Then observe $A_{\alpha}' \subseteq A_\alpha$, so $A_{\alpha}' = A_\alpha$ by maximality of $(A_\alpha)$. 
Next he shows that each $A_\alpha$ is a singleton. Hold fixed an $\alpha$, select $x_\alpha \in A_\alpha$, and put $B_\beta := A_\beta \cap f^{-1}_{\alpha\beta}(x_\alpha)$ if $\alpha \leq \beta$; otherwise put $B_\beta = A_\beta$. Bourbaki shows that $(B_\beta)_{\beta \in D}$ belongs to $\Sigma$, and clearly $B_\beta \subseteq A_\beta$ for all $\beta \in D$, so $A_\beta = B_\beta$ for all $\beta$ by maximality of $(A_\beta)$, and in particular $A_\alpha = B_\alpha = \{x_\alpha\}$. 
Finally, he shows $f_\alpha(L) = \bigcap_{\alpha \leq \beta} f_{\alpha\beta}(F_\beta)$. The direction $\subseteq$ is immediate. For the other direction, suppose $x_\alpha \in \bigcap_{\alpha \leq \beta} f_{\alpha\beta}(F_\beta)$, and put $B_\beta := f_{\alpha\beta}^{-1}(x_\alpha)$ if $\alpha \leq \beta$, else put $B_\beta = F_\beta$. One argues that $(B_\beta)_{\beta \in D}$ belongs to $\Sigma$ and that there is a maximal element $(A_\beta)$ that is $\preceq$-above $(B_\beta)$, with $A_\beta = \{y_\beta\}$. The tuple $y = (y_\beta)_{\beta \in D}$ belong to the limit $L$, and $f_\alpha(y) = y_\alpha = x_\alpha$, as was to be shown. $\Box$ 
As for question 2. of the OP, there is the slightly more general statement that if the directed set $D$ admits a countable cofinal subset and all the $f_{\alpha\beta}$ are surjective, then the limit is nonempty (this is without all the extra baggage of the $S_\alpha$ above). The countable cofinal set allows us to reduce to the case $D = \mathbb{N}$ without loss of generality, and then from there it's an easy application of dependent choice. See Proposition 5 in Bourbaki, just before the theorem we quoted above. 
A: *

*A proof is given in Bourbaki's TG.I.9.6.

*General conditions for a projective limit in the category of sets to be nonempty are given in Bourbaki's E.III.7.4.
