Let $\kappa$ denote the smallest possible size of some set $X$ with the properties you've described. I claim that $\kappa$ is just another cardinal in disguise, the unbounding number $\mathfrak b$. I'll define $\mathfrak b$ in a moment, but let me first point out some consequences of this observation:
$(1)$ $\kappa$ is uncountable, but of size at most $\mathfrak c = |\mathbb R|$. As David Gao points out in the comments, this settles the question if you assume CH, because then you must have $\kappa = \mathfrak c = \aleph_1$.
$(2)$ On the other hand, it is consistent with ZFC that $\mathfrak c$ is as large as you like, while $\kappa$ is any smaller uncountable cardinal. For example, you can have $\kappa = \aleph_{42}$ while $\mathfrak c = \aleph_{137}$, or whatever.
This question, and others like it, are used to define uncountable cardinal numbers known as cardinal characteristics of the continuum.
Claim $(1)$ follows from the known fact that $\mathfrak b$ is uncountable and $\mathfrak b \leq \mathfrak c$. Claim $(2)$ follows from the fact that it is known to be consistent to have $\mathfrak b$ be whatever you like, while $\mathfrak c$ is any larger value you like (subject to the ZFC-provable constraint that $\mathfrak b$ is a regular cardinal, and $\mathfrak c$ does not have countable cofinality).
If $f,g$ are functions $\mathbb N \rightarrow \mathbb N$, we say that $g$ dominates $f$ if $g(n) \geq f(n)$ for all sufficiently large values of $n$. The unbounding number $\mathfrak b$ is defined to be the smallest size of a family $B$ of functions $\mathbb N \rightarrow \mathbb N$ such that no single function $\mathbb N \rightarrow \mathbb N$ dominates every member of $B$. Equivalently, given any $g: \mathbb N \rightarrow \mathbb N$ (no matter how fast-growing), there is some $f \in B$ such that $f(n) > g(n)$ infinitely often.
Proof that $\mathfrak b \leq \kappa$:
Suppose $X$ is a family of sequences converging to $0$, and $|X| < \mathfrak b$. We will find an unbounded sequence $\langle y_n \rangle$ such that $\langle x_ny_n \rangle$ is bounded for every sequence $\langle x_n \rangle$ in $X$. To each sequence $\langle x_n \rangle$ in $X$, associate a function $f$ such that $f(n)$ is defined to be the least $M$ such that $x_i \notin (-1/n,1/n) \setminus \{0\}$ for any $i \leq M$. I.e., $f$ tells you how long it takes $\langle x_n \rangle$ to name a nonzero number in the $1/n$-ball around $0$. Because $|X| < \mathfrak b$, there is some function $g$ that dominates every $f$ arising in this way from a sequence in $X$. Now define $y_n$ to be a sequence of the form $1,1,\dots,1,1/2,1/2,\dots,1/2,1/3,1/3,\dots,1/3,\dots$, so that the first $1/n$ in the sequence does not occur until spot $g(n)$. This sequence $\langle y_n \rangle$ has the desired property that $\langle x_ny_n \rangle$ is bounded for every sequence $\langle x_n \rangle$ in $X$. The reason is that if $g$ dominates the corresponding $f$ after time $N$, then $x_iy_i$ is never bigger than $N$.
Proof that $\kappa \leq \mathfrak b$:
To every increasing function $f: \mathbb N \rightarrow \mathbb N$, associate a sequence $\langle x_n \rangle$ as follows. The sequence is of the form $1,1,\dots,1,1/2,1/2,\dots,1/2,1/3,1/3,\dots,1/3,\dots$, with the first 1/2 occurring in spot $f(1)$, and the first 1/3 occurring in spot $f(2)$, etc. Now let $B$ be a family of functions $\mathbb N \rightarrow \mathbb N$ as described in the definition of $\mathfrak b$, with $|B| = \mathfrak b$, and let $X$ be the corresponding family of sequences. I claim that this $X$ works. To see this, fix an arbitrary unbounded sequence $\langle y_n \rangle$. Define a function $h$ such that $h(1)$ tells you the least $n$ with $y_n > 1$, and $h(2)$ tells you the least $n$ with $y_n > 4$, and generally, $h(i)$ tells you the least $n$ with $y_n > i^2$. By our choice of the family $B$, there is some $f \in B$ that is not dominated by $h$. But then the corresponding sequence $\langle x_n \rangle$ will make the product $\langle x_ny_n \rangle$ unbounded. The reason is that we have $f(i) > h(i)$ for infinitely many values of $i$. Given some such $i$, if $f(i) = n$ then $x_ny_n > x_ni^2 \geq (1/i)i^2 = i$, because $x_n \geq 1/i$ if $f(i) > h(i)$.
