Show that $A$ is dense in $X$ if and only if every nonempty open set of $X$ contains a point of $A$. Show that $A$ is dense in $X$ if and only if every nonempty open set of $X$ contains a point of $A$.
I did not think there was much to prove for this, if using the theorem:$x \in \bar A$ if for each open set $U$ containing $x$, $U \cap A \neq \varnothing$.
Attempt:
$(\Rightarrow)$
If A is dense, then $\bar A=X$. So for each $x \in X$, there is an open set $U$ containing $x$ with $U \cap A \neq \varnothing$. Since $x$ was arbitrary this holds for each $x \in X$, so each open set in $X$ contains points in $A$.
$(\Leftarrow)$If each open set contains points in $A$, for any $x \in X$ and open set $U$ containing $x$, $U \cap A \neq \varnothing$. So $\bar A=X$.
I do not know if this proof is actually okay. I have seen a different proof in "topology without tears", that is much more involved, using a proof by contradiction in one direction. Am I over simplifying this, and is my proof unsuccessful? Any help will or criticism be appreciated.

It's indeed that simple: if $\overline{A} = X$ and $U$ is non-empty open, then we can argue as you did: let $x \in U$ then $x \in \overline{A}$ , so any open neighbourhood of $x$ intersects $A$ (in particular $U$ too) so $U \cap A \neq \emptyset$.
Conversely, if $A$ intersects every non-empty open set, then $\overline{A}=X$ or else $U = X\setminus \overline{A}$ would be a non-empty open set disjoint from $A$, or: let $x \in X$, $U$ any open neighbourhood of $x$, by assumption $A$ intersects $U$ so by the criterion you mentioned $x \in \overline{A}$, so $X \subseteq \overline{A}$ and so $X=\overline{A}$.
