Fix a ideal maximal $\mathcal{J}\subseteq 2^{\omega}$ and a countable dense subset $D\subseteq\mathcal{J}$, for every $x\in2^{\omega}$ exists a sequence $\{d_{n}\}_{n\in\omega}\subseteq D$ such that $\displaystyle{\lim_{n\rightarrow\infty}d_{n}}=x$?
The above is obvious if $x\in\mathcal{J}$, but if $x\in 2^{\omega}\setminus\mathcal{J}$ I do not know why it happens. The above is stated in the proof of Lemma 20 of https://arxiv.org/pdf/1108.2533.pdf
This holds only if $\mathcal{J}$ is nonprincipal; for instance, $\{x\in 2^\omega : x(0)=0\}$ is a closed maximal ideal and hence not dense.
When $\mathcal{J}$ is nonprincipal, it contains $\mathit{Fin}$, the family of all the finite subsets of $\omega$. Now, if $\mathcal{D}$ is dense in $\mathcal{J}$, $$ 2^\omega=\overline{\mathit{Fin}}\subseteq \overline{\mathcal{J}}=\overline{\mathcal{D}}, $$ where the last equality holds because $\mathcal{J}\subseteq \overline{\mathcal{D}}$.
Hence $\mathcal{D}$ is dense in $2^\omega$.