Let $X$ be a separable sequential space with unique sequential limits ($US$). Can we prove that $X$ has cardinality at most $\mathfrak c=2^{\aleph_0}$?
Context.
If $X$ were Fréchet-Urysohn instead of merely sequential, we could apply the argument from this answer. As it stands, I would like to have a proof of the more general result for $\pi$-base.
Suppose $D\subseteq X$ is a countable dense set. For every ordinal $\alpha$, define the set the set $D_\alpha$ by transfinite recursion as follows:
$D_0=D$.
$D_{\alpha+1}$ is the sequential limit set of $D_{\alpha}$.
$D_{\alpha}=\bigcup_{\beta<\alpha}D_\beta$ when $\alpha$ is a limit ordinal.
If $|D_\alpha|\leq \mathfrak c=2^{\aleph_0}$, then since each $x\in D_{\alpha+1}$ is a limit of some sequence $(x_i)$ of points in $D_\alpha$, we have by $US$ that
$$\left|D_{\alpha+1}\right|\leq \left|D_\alpha^{\mathbb N}\right|\leq \left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0}=\mathfrak c,$$
and if $\alpha$ is a limit ordinal, with $|D_\beta|\leq \mathfrak c$ for each $\beta<\alpha$, then $$|D_\alpha|\leq |\alpha|\cdot \mathfrak c.\tag{1}$$
If $\alpha$ is also countable we then have $|D_\alpha|\leq \aleph_0 \cdot \mathfrak c=\mathfrak c$.
Thus by transfinite induction, $\left|D_\alpha\right|\leq \mathfrak c$ for each countable ordinal $\alpha$.
But then by (1), we also have $$|D_{\omega_1}|\leq \aleph_1\cdot \mathfrak c=\mathfrak c.$$
Moreover, we claim $D_{\omega_1}$ must be sequentially closed. Indeed, if $x_i\to x$, and each $x_i\in D_{\omega_1}$, then each $x_i\in D_{\alpha_i}$ for some $\alpha_i<\omega_1$, so all the terms lie in $D_\alpha$ for $\alpha=\sup_{i\in\mathbb N}\alpha_i<\omega_1$, so we must have $x\in D_{\alpha+1}\subseteq D_{\omega_1}$.
Since $X$ is sequential, we have that $D_{\omega_1}$ is actually closed, so by density of $D$ we have $X=D_{\omega_1}$, whereby $|X|=|D_{\omega_1}|\leq \mathfrak c$.