There is a simple counterexample $\mathcal{A}$ of size continuum. To see this, observe that there are only continuum many subsequence indexing functions $\alpha:k\mapsto \alpha(k)$, and for each such $\alpha$, we may pick a sequence $\{x^\alpha_n\}_n$ in such a way that $\{x^\alpha_{\alpha(k)}\}_k$ is not convergent. That is, we define the $\alpha$th sequence specifically so that $\alpha$ does not work with it. Because of this, there can be no common subsequence indexing $\alpha$ for this family of sequences, since every $\alpha$ fails with the $\alpha$th sequence $\{x^\alpha_n\}_n$.
Beyond this, in the case that the continuum hypothesis fails, one might hope for a smaller counterexample. We can introduce a natural cardinal characteristic here, namely, the smallest size of a set $\mathcal{A}$ admitting no such common subsequence indexing. Let us call it the common convergence number, denoted 𝕔. So we've established $$\aleph_1\leq 𝕔\leq 2^{\aleph_0}.$$ Perhaps this cardinal will be provably equal to one of the other better known cardinal characteristics. The fact that subsequences of convergent sequences are also convergent seems to suggest that it could be consistent with ZFC that the smallest counterexample could be less than continuum.
In light of the counterexample, we seem led naturally to the question:
Question. How large is the smallest counterexample? In other words, how big is 𝕔?
It makes sense to me to consider 𝕔 as defined with respect to real sequences only, but I am unsure whether there is any dependence on the underlying space.
(I encourage others to post further answers about this on this same question thread.)
As a first effort in this direction, let me prove the following.
Theorem. The common convergence number is at most the splitting number $\frak{s}$. $$ 𝕔\leq \frak{s}$$ Indeed, there is a family $\mathcal{A}$ of $\frak{s}$ many binary sequences with no common convergent subsequence indexing.
Proof. Suppose that $S$ is a splitting family, which means that elements of $S$ are infinite subsets of $\mathbb{N}$, such that for every infinite set $b\subseteq\mathbb{N}$ is split by some $a\in S$, which means that $b-a$ and $b\cap a$ are both infinite. The splitting number $\frak{s}$ is the size of the smallest such splitting family.
Given $S$, let $x^a_n$ be $0$ or $1$ depending on whether $n\in a$. I claim there is no common convergent subsequence indexing $\alpha$. For any proposed $\alpha$, let $b=\text{ran}(\alpha)$. So there is some $a\in S$ with $b-a$ and $b\cap a$ both infinite. It follows that the subsequence of $\{x^a_n\}_n$ defined by $\alpha$ will not converge, since the values of $x^a_{\alpha(k)}$ will be infinitely often $0$ and infinitely often $1$. So this family is a counterexample to common convergence of size $\frak{s}$. Thus, the smallest such counterexample is at most $\frak{s}$. $\Box$
It is known that the splitting number can be $\aleph_1$, even when the continuum is large.
Let $𝕔_{\{0,1\}}$ be the common convergence number, when defined for 0/1-valued sequences only.
Theorem. $𝕔_{\{0,1\}}$ is exactly the splitting number $\frak{s}$.
Proof. The counterexample provided above shows that $𝕔_{\{0,1\}}\leq\frak{s}$. For the converse, suppose that we have a family $\mathcal{A}$ of 0/1-valued sequences, with $\mathcal{A}$ of size less than $\frak{s}$. We may regard these sequences as characteristic functions of sets, and so there is an infinite set $b\subseteq\mathbb{N}$ that is either almost contained in or almost disjoint from every set arising from $\mathcal{A}$. It follows that $b$ can be used for a common convergent subsequence indexing. So every binary family of fewer than the splitting number of sequences does admit a common convergent subsequence indexing. $\Box$
Of course, $𝕔\leq 𝕔_{\{0,1\}}$, since there might be a smaller counterexample with nonbinary sequences.