What collection of subsets $S$ of natural numbers s.t. $\{x_n\}_{n \in S}$ converges for each $S$ implies $\{x_n\}_{n \in N}$ converges? I am generalizing this question. The gist is there is a countable collection of $S \subset N$ such that $\{x_n\}_S$ converges for each $S$ but $\{x_n\}_{n=1}^\infty$ doesn't. (Order elements in $S$ as normal to make notion of sequence meaningful.)
However, it is easy to show that if for every $S \subset N$ with $S^c$ infinite we have $\{x_n\}_{n \in S}$ convergent, then $\{x_n\}_{n \in N}$ converges. 
(Take some $S$. Then both $\{a_n\}_{n=1}^\infty = \{x_n\}_{n \in S}$ and $\{b_n\}_{n=1}^\infty = \{x_n\}_{n \in S^c}$ converge to $a$ and $b$ respectively. Then take $a_{2n}$ and $b_{2n}$, zip them together as a subsequence, and get that they converge to some $c$ (realize the complement is infinite). Hence $$a = \lim a_n = \lim a_{2n} = c =  \lim b_{2n} = \lim b_n = b.$$ It follows $x_n$ converges because $a_n$ and $b_n$ zipped together compose $x_n$ completely.)
You can see that the condition required is more than needed. We only need some $S$, with $S$, $S^c$ giving convergence, and some $T \subset S$ and $U \subset S^c$ with $S$ and $U$ infinite and such that $S \cup T$ gives convergence.
So originally I was wondering if it is necessary to have a uncountable collection of $S$ such that $\{x_n\}_S$ converges to get that $\{x_n\}_{n=1}^\infty$ converges. But this is not the case. In fact, you can get away with just a finite number of such $S$.
My Question: 
Does there exist a countable collection of infinite subsets of $N$, $\{S_m\}_{m=1}^\infty$, such that (a) $S_i^c$ is infinite and not in the collection, (b) furthermore $(S_i \cup S_j)^c$ is never finite, and (c) $\{x_n\}_{n \in S_m}$ converges for each $S_m$ implies $\{x_n\}_{n=1}^\infty$ converges.
Note these conditions are satisfied in the motivation question but the conditions discussed above where we do have an implication are destroyed. The more general question is what are the conditions for a collection of subsequences converging imply the original sequence converges, but that is likely a very hard question.
 A: Given your choice of tags on this question and the argument you gave in the third paragraph, I will assume we are working in a context in which limits of sequences are unique, such as metric spaces or more generally Hausdorff spaces.  Under that assumption, you can get such a collection with just four subsets.
Partition $\mathbb{N}$ into five infinite subsets $A$, $B$, $C$, $D$, and $E$.  Let $S_1=A\cup B$, let $S_2=B\cup C$, let $S_3=C\cup D$, and let $S_4=D\cup E$.  It is easy to verify that these satisfy your conditions (a) and (b).  Now suppose you have a sequence $(x_n)$ which converges when restricted to $S_m$ for $m=1,2,3,4$; let $a_m$ be the corresponding limit.  Since $S_1$ and $S_2$ contain $B$ as a common subsequence, we must have $a_1=a_2$.  Similarly we have $a_2=a_3$ and $a_3=a_4$, so actually all the limits are the same, which we can call $a$.  It follows that the entire sequence must converge to $a$.
However, there is an obvious strengthening of your axioms (a) and (b) which makes there be no examples.  Namely, you should require that the complement of any finite union of the $S_m$ to be infinite, not just binary unions.  Suppose $\{S_m\}$ satisfies this condition.  Define a set $T=\{t_n\}$ by induction by letting $t_n$ be the least element of $(S_1\cup\dots\cup S_n)^c$ which is not equal to $t_k$ for any $k<n$.  Then $T$ is infinite, and has the property that $T\cap S_m$ is finite for all $m$ (since $t_n\not\in S_m$ for $n\geq m$).  Now consider a sequence $(x_n)$ which has one constant value $a$ on $T$ and a different constant value $b$ on $T^c$.  Since $T\cap S_m$ is finite, the restriction of this sequence to each $S_m$ is eventually constant with value $b$, and so converges to $b$.  But the sequence itself does not converge, since it is equal to both $a$ and $b$ infinitely often.
