TITLE: Show $\Bbb R ^n$ with a finite set removed is still connected.
QUESTION [5 upvotes]: The question is:

Show that if $A⊆\Bbb R^n$ is finite and $n≥2$, then the set $\Bbb R^n\backslash A$ is connected in $(\Bbb R^n,τ_d)$ where $τ_d$ is the topology induced by the Euclidean distance in $\Bbb R^n$.

My attempt:
Denote $A^C=\Bbb R^n\backslash A$ and suppose $A=\{a_1,…,a_n\}$. Assume by contradiction that $A^C$ is not connected, then there exists $O_1,O_2∈τ_{A^C }$ st. $A^C=O_1⋃O_2,O_1⋂O_2=∅$. Here $\tau_{A^C}$ is a subspace topology.
Intuitively, part of $A$, denoted by $A_1$ is "enclosed" in $O_1$ and the rest, denoted by $A_2$ are "enclosed" in $O_2$. Then $G_1=O_1\bigcup A_1$ is an open set in $\tau_d$, and $G_2=O_2\bigcup A_2$ is also an open set in $\tau_d$, and $\Bbb R^n=G_1\bigcup G_2$. However, $G_1, G_2$are disjoint, contradicting with the fact that $\Bbb R^n$ is connected.
However I got stuck when trying to implement this idea. Can anyone provide some help? Thank you!

REPLY [5 votes]: Let's show that $\mathbb{R}^n \setminus A$ is path-connected. Take any two points $x$ and $y$. One would like to exhibit a path from $x$ to $y$. It suffices to show that there is at least one point $p$ so that $x$ and $y$ can both be connected to $p$ via a straight line that misses all the points in $A$. 
For each point $a \in A$, let $L_{a}$ be the line passing through $x$ and $a$, and let $N_{a}$ be the line passing through $y$ and $a$. This describes finitely many lines. $\mathbb{R}^n$ is not a finite union of lines, because $\mathbb{R}^n$ intersects the sphere of radius $1$ at infinitely many points (the whole sphere), but a line can only intersect a sphere at $2$ points, and hence finitely many lines can only intersect a sphere at finitely many points. Thus there is some point $p$ that is not contained in the union of these lines. Call this point $p$. Can you see why this $p$ works?
[This proof generalizes nicely to the situation when $A$ is countably infinite]