Suppose every point x in an ordered set X is the limit of some strictly monotone sequence. Is X uncountable?

Question

If have an ordered set $X$ and for each $x\in X$ I can construct a sequence $(x_n)$ as follows: (1) $x_n\rightarrow x$ and (2) $x_n$ is strictly monotone (lets say, increasing, for concreteness). My intuition tells me that $X$ has to be uncountable but I can't seem to prove it. Is this true or is my intuition wrong? Note: the topology I use in $X$ is the topology of intervals -- i.e the topology with basis given by all sets of the form $\{x: a>x>b\}$, where $a,b\in X$. My intuition is as follows: the property I described implies that I can associate to each $x$ a sequence of (countably many, distinct) points of $X$. This should imply that the cardinality of $X$ is at least $\mathbb{N}^{\mathbb{N}}$, right?

Answer 1

Every dense set satisfies the property.

You can build a sequence that converges to $x$ recursively by first taking a point $a_1\in (-\infty,x)$ and then making $a_i$ be in $(a_{i-1},x)$.

It makes sense that countable dense sets exist because $\mathbb R$ has a countable basis (the intervals with rational end-points).