A collection of sequences that cannot all be made to converge I am trying, mostly out of curiosity, to exhibit an infinite countable set $X$ of real, null sequences, such that given a sequence of signs $(s_n)\in\{-1,1\}^{\mathbb{N}},$ at least one of $(x_n)\in X$ will lead $\sum s_nx_n$ to diverge.
Every construction I have tried has failed, yet I am convinced such an $X$ should exist. Would appreciate any clever ideas, or a proof of the contrary.
Source: my question is inspired by this one, which treats the case $X$ finite: in that particular case there is no such set because it is possible to construct $(s_n)$ so that every $(x_n)\in X$ leads to $\sum s_nx_n$ convergent. The case where $X$ has cardinality $|\mathbb{R}|$ is almost trivial, so I am wondering what happens when $X$ is merely countable.
 A: Partial answer:  If all sequences in the countable collection $\mathcal{X}$ are square-summable then almost all $\{S_n\}_{n=1}^{\infty}$ will make all modified sums converge.   
Proof: Take any real-valued sequence $\{x_n\}_{n=1}^{\infty}$ that satisfies $\sum_{n=1}^{\infty} x_n^2 < \infty$. Now form $\{S_n\}$ according to a random i.i.d. sequence with $Pr[S_n=1]=Pr[S_n=-1]=1/2$.  Define the modified sum $M_n = \sum_{i=1}^n x_i S_i$. Then, the sequence $\{M_n \}_{n=1}^{\infty}$ is a martingale and for all $n$:
$$ E[M_n^2] = \sum_{i=1}^n x_i^2  \leq \sum_{i=1}^{\infty} x_i^2 < \infty$$
So $M_n$ converges with probability 1 by the (quadratic) martingale convergence theorem.
Now let $\mathcal{X}=\{\{x_n[k]\}_{n=1}^{\infty}\}_{k=1}^{\infty}$ be a countably infinite collection of square summable sequences.  Choose $\{S_n\}$ randomly as before and define $\{M_n[1], M_n[2], M_n[3],...\}$ as the corresponding martingale for each sequence in the collection, so:
$$ M_n[k] = \sum_{i=1}^n x_i[k]S_i $$
Then: 
$$Pr[\mbox{$M_n[k]$ converges for all $k$}]  \geq 1-\sum_{k=1}^{\infty}Pr[\mbox{$M_n[k]$ does not convege}] = 1 $$
So, with prob 1, the randomly chosen sequence $\{S_n\}$ makes them all converge. This of course also means there exists an $\{S_n\}$ sequence that makes them all converge. $\Box$

Possible idea for general case:  If we know we can always get a sign-sequence $\{s_n\}$  that makes all null sequences in a finite collection converge, what about doing something along the lines of defining $\{s_n[k]\}$ as the sign-sequence that makes the first $k$ null sequences converge, and then constructing $\{s_n^{new}\}$ from $\{s_n[k]\}$ in some Cantor-diagonal-like way.  
A: This is a writeup of the general case along the lines I suggested in my first answer (I am actually posting two answers since they are qualitatively different). 
Proposition: There is always a sign-sequence $\{s_n\}$ that makes all modified sums of the countably infinite collection of null sequences converge. 
Proof construction: 
Let $\{x_n[k]\}_{n=1}^{\infty}$ be a countably infinite collection of null sequences indexed by $k \in  \{1, 2, 3, …\}$.  For each positive integer $m$, let $\{s_n[m]\}_{n=1}^{\infty}$ be a sign-sequence (consisting of 1 and -1 values) such that $\sum_{i=1}^{n} x_i[k]s_i[m]$ converges to a finite value (as $n\rightarrow\infty$)  for all $k \in \{1, …, m\}$ (we know this is possible by the link given by the asker of this question).  
Note that a real-valued sequence will converge if and only if it is a Cauchy sequence. So, for every $\epsilon>0$ and every positive integer $m$, there is a positive integer $R_m(\epsilon)$ such that if $a, b$ are integers that are greater than or equal to $R_m(\epsilon)$ then: 
$$\left|\sum_{i=1}^a x_i[k]s_i[m] - \sum_{i=1}^b x_i[k]s_i[m]\right| \leq \epsilon \quad \forall k \in \{1, ..., m\} $$  
Now form $\{s_n^{new}\}$ as follows: 
-Define $s_n^{new}=s_n[1]$ for $n \in \{1, …,G_1\}$, where $G_1 = R_2(2^{-1})$.  
-Define $s_n^{new} = s_n[2]$ for $n \in \{G_1+1, …, G_2\}$, where $G_2 = max[G_1+1,R_3(2^{-2})]$.  
-Define $s_n^{new} = s_n[3]$ for $n \in \{G_2+1, …, G_3\}$, where $G_3 = max[G_2+1, R_4(2^{-3})]$. 
and so on, so: 
$s_n^{new} = s_n[m]$ for $n \in \{G_{m-1}+1, …, G_m\}$, where $G_m = max[G_{m-1}+1, R_{m+1}(2^{-m})]$.  
Now fix a positive integer $k$. The following claim establishes the proposition. 
Claim: The sequence $\sum_{i=1}^n x_i[k]s^{new}_i$ is Cauchy. 
Proof of claim: Fix a positive integer $m$ and let $a,b$ be integers larger than $G_m$, with $a<b$. We want to show that the following value vanishes as $m\rightarrow \infty$: 
$$ \left|  \sum_{i=1}^a x_i[k]s^{new}_i - \sum_{i=1}^b x_i[k]s^{new}_i \right| $$
Now if $a,b$ are in the same interval $\{G_{r-1}+1, .., G_{r}\}$ for some $r$, then the difference is at most $2^{-(r-1)}$, while if they are in different intervals the difference is at most a sum of geometrically decreasing terms, which is still small. $\Box$
