Dense subset in product topology

Question

Suppose $D\subset R^2$ (with the product topology in $R\times R$ ) is dense (in $R$) and also connected. Must $D = R^2$? Explain with proof or example.

Answer 1

If $A$ is countable in the plane, $D=\Bbb R^2\setminus A$ is (path-)connected: there are uncountably many disjoint(-except-endpoints) paths between any two points in $\Bbb R^2$ and removing $A$ can only kill countably many such paths. So $D$ is path-connected.

$D$ is also dense, as any non-empty open set in $\Bbb R^2$ is uncountable, so remains that way (in particular, non-empty) after cutting out $A$.

Answer 2

No, let $D$ be the union of all lines $y=kx$ with rational slope $k$.