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.
Let $(X,d)$ be a metric space and let $E$ be a dense subset of $X.$ Let $B=\{B_d(x,q): x\in E\land q\in \mathbb Q^+\}.$ Then $B$ is a base for $X.$
Proof: Let $p\in U$ where $U$ is open in $X.$ It suffices to show that $p\in b\subset U$ for some $b\in B.$
There exists $r>0$ such that $B_d(p,r)\subset U,$ and there exists $q\in (0,r/2)\cap \mathbb Q^+.$ And there exists $e\in E$ with $d(p,e)<q/2$ because $E$ is dense in $X.$ So $p\in B_d(e,q)\in B.$
And $y\in B_d(e,q)\implies d(y,p)$ $\leq d(y,e)+d(e,p)<q+q/2<2q<r$ $\implies y\in B_d(p,r)\subset U.$ So $p\in B_d(e,q)\subset U.$.... Q.E.D.
If $E$ is countable then $B$ is countable. When $X=\mathbb R^n$ with the usual topology, let $E$ be the set of points with rational co-ordinates. Then $E$ is dense and countable so $B$ is a countable base for $\mathbb R^n.$
EDIT: formatting second paragraph in proof
Best Answer
Consider $f(x) = 0$ and corresponding set $G_0$. Then $$ \text{Int}\ G_0 = \{\textit{g: g continuous and changes sign somewhere}\} \\ \overline{G_0} = \{\textit{g: g continuous and not separated from 0, i.e. }\inf|g| = 0\} $$ It is not meagre as its interior is not empty. When it comes to size, $G_0$ is what remains in $C(\mathbb{R})$ when you remove all functions of constant nonzero sign.
For arbitrary $f$ you have $G_f = f + G_0$.