This answer is the proof given by Ashutosh, but formulated in terms of the splitting number.

**Proposition**
If the splitting number $s$ is $\aleph_{1}$, then every nonseparable metric space contains a sequence of subsets with no convergent subsequence.

Proof: Following Sierpinski, since the metric space $M$ is non-separable, there exist $d > 0$ and a sequence $\{p_\xi\}_{\xi<\omega_{1}}$ of points in $M$ such that $\varrho(p_\xi,p_\eta)\ge d$ for $\xi<\eta<\omega_{1}$, where $\varrho(x,y)$ is the metric on $M$.

Let $S$ be a splitting family (for $[\omega]^{\omega}$) of size $\aleph_{1}$, $S = \lbrace s^\xi : \xi < \omega_{1} \rbrace$, where $s^\xi = \langle n_1^\xi,n_2^\xi,n_3^\xi,\ldots\rangle$; for a given $k \in \mathbb{N}$, let $E_k$ be the set of all $p_\xi$ such that $k\in \{n_1^\xi,n_2^\xi,\ldots\}$.

The sequence $E_1,E_2,E_3,\ldots$ does not contain any convergent subsequence. For, if $E_{k_1}, E_{k_2},\ldots$ where $k_1<k_2<\cdots$ is an arbitrary subsequence of $E_1,E_2,\ldots$, then there exists $\alpha<\omega_{1}$, such that $s^\alpha$ splits $K = \lbrace k_{n} : n < \omega \rbrace$ into infinite $a = K \cap s^{\alpha}$ and $b = K \setminus s^{\alpha}$ say. Now $p_\alpha\in E_{k_{i}}$ for $k_{i} \in a$ and $p_\alpha \notin E_{k_{j}}$ for $k_{j} \in b.$ The open ball with centre $p_\alpha$ and radius $d$ (which is actually just the singleton $p_\alpha$) intersects each $E_{k_{i}}$ for $k_{i} \in a$ non-trivially but is disjoint from $E_{k_{j}}$ for $k_{j} \in b$. Consequently, $E_{k_1},E_{k_2},E_{k_3},\ldots$ is not convergent. So the sequence $E_1,E_2,\ldots$ contains no convergent subsequence, q.e.d.

**Corollary** (Sierpinski)
CH implies the assertion (*) every nonseparable metric space contains a sequence of subsets with no convergent subsequence.

Proof. CH implies $s = \aleph_{1}$.

**Corollary** (Ashutosh)
The assertion (*) does not imply CH.

Proof. It is relatively consistent that $s = \aleph_{1} < 2^{\aleph_{0}}$. q.e.d.