Show two notions of dense are equivalent

This question follows from another one Topology proof: dense sets and no trivial intersection

Show that given a topological space $(X, \mathcal{T}), D \subseteq X$

Then $D$ is dense iff $\forall U \in \mathcal{T}, U \neq \varnothing, D \cap U \neq \varnothing$

Here my try:

$(\Rightarrow)$ $D$ is dense if $\overline D = X$. By definition, $\overline{D} = \{x \in X| \forall U \in \mathcal{T}, x \in U \implies U \cap D \neq \varnothing\}$ so $\overline{\overline D} = \{x \in X| \forall U \in \mathcal{T}, x \in U \implies U \cap \overline D \neq \varnothing\}$

Since $\overline D = X$, and $U \subseteq X$, therefore $\forall x \in X, \forall U \in \mathcal{T}, x \in U \implies U \cap \overline D \neq \varnothing$

But $\overline D = \overline{\overline D}$, therefore $\forall x \in X, \forall U \in \mathcal{T}, x \in U \implies U \cap D \neq \varnothing \Leftrightarrow \forall U \in \mathcal{T}, U \neq \varnothing, D \cap U \neq \varnothing$

($\Leftarrow$) Given $D \subseteq X$, $\forall U \in \mathcal{T}, U \subseteq X, D \cap U \neq \varnothing$, we want to show that $\overline D = X$. I think this proof is immediate but am I not sure how to show this...

Can someone check my attempt and show me how to continue with $\Leftarrow$

Your first proof is correct but unnecessarily complicated. Suppose that $\operatorname{cl}D=X$, and let $U$ be any non-empty open set in $X$. $U\ne\varnothing$, so there is an $x\in U$; and $x\in X=\operatorname{cl}D$, so $U\cap D\ne\varnothing$.
For the other direction assume that each non-empty open subset of $X$ meets $D$, and let $x\in X$ be arbitrary. If $U$ is any open nbhd of $x$, then $U\cap D\ne\varnothing$, so $x\in\operatorname{cl}D$. Thus, $X\subseteq\operatorname{cl}D\subseteq X$, so $\operatorname{cl}D=X$.