Given a dense subset of $\mathbb{R}^n$, can we find a line that intersects it in a dense set? I have some difficulties in the following question.

Let $S$ be a dense subset in $\mathbb{R}^n$. Can we find a straight line $L\subset\mathbb{R}^n$ such that $S\cap L$ is a dense subset of $L$.

Note. From the couterexample of Brian M. Scott, I would like to ask more. If we suppose that $S$ has a full of measure. Could we find a straight line $L$ such that $S\cap L$ is a dense subset of $L$?

For each $n>1$ it’s possible to construct a dense subset $D$ of $\Bbb R^n$ such that every straight line in $\Bbb R^n$ intersects $D$ in at most two points.
Let $\mathscr{B}=\{B_n:n\in\omega\}$ be a countable base for $\Bbb R^n$. Construct a set $D=\{x_n:n\in\omega\}\subseteq\Bbb R^n$ recursively as follows. Given $n\in\omega$ and the points $x_k$ for $k<n$, no three of which are collinear, observe that there are only finitely many straight lines containing two points of $\{x_k:k<n\}$; let their union be $L_n$. $L_n\cup\{x_k:k<n\}$ is a closed, nowhere dense set in $\Bbb R^n$, so it does not contain the open set $B_n$, and we may choose $x_n\in B_n\setminus\big(L_n\cup\{x_k:k<n\}\big)$. 
Clearly $D$, so constructed, is dense in $\Bbb R^n$, since it meets every member of the base $\mathscr{B}$, and by construction no three points of $D$ are collinear.
