Question: let denote $U$ be unit square in complex plane, whose corners are $0, 1, i,$ and $1+i$. then, the set
$S= \{a+bi: a, b\text{ are rationals inside }U\}$ is connected?
My attempt: clearly $S$ is not polygonally connected! (Since we can find path between any two points in $S$ which contains the point which do not belongs to $S$) but this does not show $S$ is connected or not!
How to show $S$ is connected or not? to show not connected: we have to show $S$ can be expressed as disjoint union of two nonempty open subsets of $S$, but I am unable to express $S$ like this! :-( So is $S$ is connected? But how? Please help me, I am stuck on this
I don't know answering our own question is fine or not. But, I think it is best way to know ,"what we get from others helpful comments". @Saulspatz sir, and @астонвіллаолофмэллбэрг Sir, please check is this is correct.
define $U_1=\{a+bi: a<\frac{1}{\sqrt{2}}\}$ and $U_2=\{a+bi: a>\frac{1}{\sqrt{2}}\}$
both are open sets in $\mathbb{C}$ and hence, by definition of subspace topology, $S∩U_1$ and $S∩U_2$ are open sets in $S$ and clearly $S= S∩U_1∪S∩U_2$ and so $S$ is written as disjoint union of nonempty open sets and hence $S$ is disconnected.