Is the unit square with dictionary ordering second countable?

Question

I'm conflicted: If we consider the set $\{x\} \times (0,1)$, for $ x \in [0,1]$, these are open in the unit square, uncountable and disjoint, but what about open intervals of the form ((a,b), (c,d)) with a,b,c,d in $ [0,1] \cap \Bbb{Q}$ and $a < c$ these are countable and cover the unit square with the lexicographical order.

Answer 1

Your first observation is the relevant one: you have an uncountable collection of pairwise disjoint, non-empty open sets, so the space cannot possibly be second countable, as each of those open sets must contain a different element of any base for the topology. Your second observation is irrelevant: yes, you have a countable open cover, but it certainly isn’t a base for the topology: none of the open sets $\{x\}\times(0,1)$ contains any member of that cover.