Assume that $E$ and $B$ are subsets of a topological space $X$. By definition the set $E$ is dense in $B$ iff the set $E\cap B$ is dense in the subspace $B$ iff $\operatorname{cl}_B(E\cap B)=B$. Using the equality $\operatorname{cl}_B(E\cap B)=\overline{E\cap B}\cap B$ and elementary observations we obtain the equivalences
$$\operatorname{cl}_B(E\cap B)=B\ \Leftrightarrow\ \overline{E\cap B}\cap B=B\ \Leftrightarrow\ \ B\subseteq\overline{E\cap B}\cap B\ \Leftrightarrow\ B\subseteq\overline{E\cap B}\ \Leftrightarrow\ B\subseteq\bar{E}.$$
Hence $E$ is dense in $B$ iff $B\subseteq\bar{E}$.
By definition $E$ is nowhere dense in $X$ iff there is no nonempty open subset (equivalently neighborhood) of $X$ in which $E$ is dense. Using the observations above, $E$ is nowhere dense in $X$ iff $\bar{E}$ does not contain any nonempty open subset of $X$ iff $\operatorname{int}\bar{E}=\emptyset$.
Indeed, in the metric setting, $E$ is nowhere dense iff there is no ball in which $E$ is dense, since any nonempty open subset of a metric space contains a ball.
In any metric space the condition of a subset being dense in every ball is equivalent to being dense in the space, because every point of a metric space is contained in a ball. Thus any dense subset of $\mathbb{R}$, for example $\mathbb{R}$, $\mathbb{R}\setminus\{0\}$ or $\mathbb{Q}$, is an example to your first question. Regarding your second question, the ball $(-1,1)\subseteq\mathbb{R}$ is dense in itself but not dense in any ball that is not contained in $(-1,1)$.
"$E$ is dense in a ball" does not mean "$\bar{E}$ is a ball". For example $(-1,1)\cup\{2\}$ is dense in the ball $(-1,1)$ but $\bar{E}=[-1,1]\cup\{2\}$ is not a ball. However, "$\bar{E}$ is a ball" implies that "$E$ is dense in a ball": if $\bar{E}$ is a ball, then $E$ is dense in the ball $\bar{E}$.
"$E$ is dense in a ball" means "$\bar{E}$ contains a ball", since according to our observations above $E$ is dense in a set $B$ iff $B\subseteq\bar{E}$.
"$E$ is dense in a ball" does not mean "$\bar{E}$ is contained in a ball". For example $\mathbb{R}$ is dense in the ball $(-1,1)$ but no ball contains its closure. On the other hand $\bar{\emptyset}=\emptyset$ is contained in every ball but $\emptyset$ is dense in no ball.
so to recap: the more modern definition
1) $X$ is a Baire space iff every countable union of closed nowhere dense sets has empty interior.
Equivalent by taking complements (note that a set $A$ is nowhere dense iff its complement $X \setminus A$ contains an open dense subset) to my favourite formulation, which seems to be more commonly used among topologists:
1') $X$ is a Baire space iff every countable intersection of open and dense subsets is dense.
(note that in any space a finite intersection of open and dense subsets is open and dense, so the countable intersection is the first "interesting" question, in a way.)
And what they call the historical definition:
2) Every non-empty open subset of $X$ is of second category.
The article call it historical because it uses a notion "category" of a subset (a subset is either first category or second category, and not both, by definition), which has fallen in some disuse. Nowhere dense sets and meagre sets (the countable unions of nowhere dense subsets) are still normal usage. Note that a first category subset is now called meagre, and the notion of "second category" is not used as much (but still occurs), so it's good to know it. But definition 1) and 2) are easily proved to be equivalent, so they give rise to the same spaces being called Baire. So we have a trivial reformulation of the "classical" definition 2 as:
2') Every non-empty open subset of $X$ is non-meagre.
Or, stated more "positively"
2'') Every meagre set has empty interior.
(otherwise the non-empty interior is a subset of a meagre set, and thus meagre etc.)
which brings us back to definition 1) again.
It's just that the Wikipedians do not like the category terminology (because it might confuse people with category theory as a branch of maths) and so choose to reformulate everything using meagre and non-meagre instead.
Best Answer
Let $A$ be a nowhere dense set. Then $(\overline{A})^o=\emptyset$. This is equivalent to saying that $(\overline{A})^c$ is dense in $X$. Let $A$ and $B$ be two nowhere dense sets. Let $S=A \cup B$. To show $S$ is nowhere dense we will show that $(\overline{S})^c$ is dense in $X$, that is $(\overline{S})^c$ meets every non-empty open set. Let $G$ be a non-empty open set. Now as $A , B$ are nowhere dense, $G\cap(\overline{A})^c\neq\emptyset$ and $G\cap(\overline{B})^c\neq\emptyset$. Also $(\overline{A})^c$,$(\overline{B})^c$ are open. Hence $G\cap(\overline{A})^c\cap(\overline{B})^c\neq\emptyset$, since $G\cap(\overline{A})^c$ is non-empty open and $(\overline{B})^c$ is dense in $X$. Thus $G\cap(\overline{A}\cup\overline{B})^c\neq\emptyset$, which implies $G\cap(\overline{S})^c\neq\emptyset$. Hence $(\overline{S})^c$ is dense in $X$. Equivalently $S$ is nowhere dense.