Define $C$ to be a set of countable number of lines. Let $p,q$ be a couple of points in $\mathbb{R}^{n}\setminus C$. Then if the line segment $\bar{pq}\cap C=\emptyset$, we're done. If $\bar{pq}\cap C\neq\emptyset$, the intersection is a set of countable number of points. Since $n>2$, at each point in the intersection, there exists an arc that goes around the point. Thus, there exists a path connecting $p$ and $q$ that goes along the line segment $\bar{pq}$ with arcs replacing the points in $\bar{pq}\cap C$. Therefore, $\mathbb{R}^{n}\setminus C$ is path-connected.
It's pointed out to me that this proof is not exactly correct. I don't know where it went wrong.
Reduce the problem to showing that $\mathbb{R}^2$ minus a countable set of points is path-connected, using one of these methods.
Let $S^2 \subset \mathbb{R}^n$ be a 2-sphere that contains $p$ and $q$ on the surface. A line intersects a sphere at most twice, so $C \cap S^2$ is countable. Remove any point $x \in S^2 \setminus (C \cup \{p, q\})$ and identify $S^2 \setminus \{x\} \cong \mathbb{R}^2$ by homeomorphism, and look for a path in $S^2 \setminus (C \cup \{x\})$.
Take the uncountable family $\mathcal{P}$ of 2-planes that contain $p$, $q$, and one point on a fixed line not coplanar with the line $pq$. $C$ is a countable set of lines, and $\mathcal{P}$ is pairwise disjoint except at $pq$ (which is not in $C$), so some $P \in \mathcal{P}$ contains no line in $C$. $P \setminus C$ is a countable set of points, as every line in $C$ intersects $P$ at most once. We'll find a path within $P$.
Now let $C \subset \mathbb{R}^2$ be a countable set of points, and $P, Q \in \mathbb{R}^2 \setminus C$ be arbitrary. There is an uncountable family of paths in $\mathbb{R}^2$ that are all disjoint except at $P$ and $Q$ themselves. (For one such family, for each point $B$ on the perpendicular bisector of $PQ$, take the circular arc $PBQ$.) Only a countable number of such paths can contain a point of $C$, so some path avoids $C$.