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.
The set $\Bbb Z$ consists entirely of isolated points, so it has no limit points. Hence, its closure is $\Bbb Z$.
Also because $\overline{\Bbb Z}=\Bbb Z$ consists of isolated points, it has no interior points, meaning its interior is empty.
To see any $k \in \Bbb Z$ is an isolated point, note that the only integer in the open interval $\left( k- \frac 12, k+\frac 12 \right)$ is $k$.
Best Answer
Nowhere dense is a strengthening of the condition "not dense" (every nowhere dense set is not dense, but the converse is false). Another definition of nowhere dense that might be helpful in gaining an intuition is that a set $S \subset X$ is nowhere dense set in $X$ if and only if it is not dense in any non-empty open subset of $X$ (with the subset topology).
For example, $\mathbb{Z}$ is nowhere dense in $\mathbb{R}$ because it is its own closure, and it does not contain any open intervals (i.e. there is no $(a, b)$ s.t. $(a, b) \subset \mathbb{\bar{Z}} = \mathbb{Z}$. An example of a set which is not dense, but which fails to be nowhere dense would be $\{x \in \mathbb{Q} \; | \; 0 < x < 1 \}$. Its closure is $[0, 1]$, which contains the open interval $(0, 1)$. Using the alternate definition, you can note that the set is dense in $(0, 1) \subset \mathbb{R}$.
An example of a set which is not closed but is still nowhere dense is $\{\frac{1}{n} \; | \; n \in \mathbb{N}\}$. It has one limit point which is not in the set (namely $0$), but its closure is still nowhere dense because no open intervals fit within $\{\frac{1}{n} \; | \; n \in \mathbb{N}\} \cup \{0\}$.