Prove that topological space $ X= {[0,1]}^{2} $ with dictionary order topology is not second countable.
I would assume that it is second countable. So then there would exist a basis $B$ of $X$ which is finite or countable. Then I would take an arbitrary $ y \in [0,1]$. Now I would look at the set $[0,1]\times\{y\} $. Because $B$ is base for each $ x \in [0,1]$ there exists $ B_{x} \in B$ so that $(x, y) \in B_{x} $ and $ B_{x} $ is a subset of $[(x, 0),(x,1)] = \{x\} \times [0, 1]$. Now I would define a function $ f \colon [0,1]\times\{y\} \to B $ so that $ f(x, y) = B_{x} $. $f$ is an injection so $B$ has to be uncountable because $[0,1]\times\{y\}$ is uncountable. Is this ok?
Your proof is fine, but maybe it's better just to note that $\{\{x\} \times (0,1):x\in X\}$ is an uncountable collection of disjoint open sets.