$\mathbb R^3$ minus a line is connected.

First: it is not reasonable to prove this with homology. Let's start with a story.

Suppose $X$ is a subset of the sphere homeomorphic to $D^k$. Then Hatcher proves in proposition 2B.1 that $S^n \setminus X$ has trivial reduced homology. (He also proves that if $X \cong S^k$, then $S^n \setminus X$ has the same reduced homology as $S^{n-k-1}$. This won't be as important to us.) The way the proof goes, roughly, is to use Mayer-Vietoris on the complement of $I^{k-1} \times [0,1/2]$ and $I^{k-1} \times [1/2, 1]$. If the homology isn't what we said above, we show the same is true for one of these sub-intervals, keep making the sub-intervals smaller, and then show that the homology of these complements ultimately reduces to the homology of $I^{k-1} \times \{p\}$ (all, again, a sketch - see the actual proof.)

It was essential that we used $I$ here, or else we wouldn't have access to the Mayer-Vietoris sequence - the complement of these sets is open! The same is not true for a copy of $\Bbb R$. If the $\Bbb R$ had closed image, you'd be able to access the problem with similar techniques (most easily by taking the one-point-compactification and thinking of this as the complement of an $S^1$ in $S^3$ and using the above technique). You might want to say "OK, well there they took a nested intersection of intervals and looked at their complement. We know the result works for embeddings of $[0,1]$. Why don't we try to take a nested union instead?" This does not work - just think about the union of discs on the $xy$-plane of increasing radius! Clearly their complements are path-connected but their increasing union is not. So there's something special going on about the 1-dimensionality of the line. (Note, by the way, that FINITE unions are easy to deal with via the Mayer-Vietoris sequence. Indeed it's not hard to show using the above facts that the complement of any $n-2$-dimensional finite simplicial complex embedded in $\Bbb R^n$ is path-connected.) Even the most souped-up duality theorems, like Alexander duality in Cech cohomology, need to assume compactness (even if they can drop local contractibility!) After all, consider the embedding of a copy of $\Bbb R$ inside the 2-sphere that winds around an annulus, with limit set $|z| = 1$ and $|z| = 3$. Then it has three path-components, which no form of duality could expect.

OK, so how do we deal with this? We, like your friend said, use dimension theory. I don't know this stuff very well, so I'm going to cite it away. (I don't have access to Hurewicz-Wallman right now, but when I do, I'll offer a more precise reference.) This MO comment says that the space of paths from $x$ to $y$ not in some compact 1-dimensional subspace of $\Bbb R^3$ is open and dense in the space of paths from $x$ to $y$, and this answer implies that the image of any unit interval in $\Bbb R^3$ is 1-dimensional. Now apply the Baire Category theorem to the subset $U_n \subset \mathcal P(x,y)$, where $U_n$ is the set of maps that miss $[n,n+1]$ in your embedding of $\Bbb R$. Because $\mathcal P(x,y)$ is a complete metric space with the supremum metric, it is a Baire space, and hence $\bigcap_{n \in \Bbb Z} U_n$ is nonempty (and, in fact, dense). Thus we conclude.

It is sufficient to show that $X=\Bbb R^3\setminus S$ is path-connected. Now for any couple of points $p,q$ in $X$ let $\overline{pq}$ be the segment joining them, then this is a path that. If $\overline{pq}\cap S=\emptyset$, there is no problem. In the case that $\overline{pq}\cap S\neq \emptyset$, take a third point $r$, different to the previous and such that $\overline{pr}\cap S=\emptyset$ and $\overline{rq}\cap S=\emptyset$, and then $\overline{pr}\cup\overline{rq}$ is a path joining $p$ with $q$.