Path-connectedness of the complement of a finite number of points Let $X=\mathbb{R}^3\setminus\{x_1,...,x_n\}$, the complement of a finite number of points. I want to show that $X$ is path-connected. I know how to do this intuitively. Let $y,z\in X$, then $t y+(1-t)z$ for $t\in [0,1]$ is either contained in $X$, and we are done, or there exists a $t_i\in [0,1]$ such that $t_i+(1-t_i)z=x_i$ for some $1\leq i\leq n$. Then, I want to make a little detour around $x_i$. How can I do this formally?
 A: Let's write formally your idea.
Let $x,y \in X$. Consider the $2n$ affine lines $L_1, ..., L_n, L'_1, ..., L'_n$, defined by the following : for $i \in \lbrace 1, ..., n \rbrace$, $L_i$ is the line that contains $x$ and $x_i$, and $L_i'$ is the line that contains $y$ and $x_i$.
Choose a point $z$ such that for all $i \in \lbrace 1, ..., n \rbrace$, $z \notin L_i$ and $z \notin L'_i$. Of course, such a point exists because a finite union of affine lines can never fill $\mathbb{R}^3$. Consider the path : $p_z : [0,1] \rightarrow X$ defined by
$$ p_z(t) = 2tz + (1-2t)x \text{ if } 0 \leq t < \frac{1}{2} \quad \text{ and } \quad p_z(t) = (2t-1)y + (2-2t)z \text{ if } \frac{1}{2} \leq t \leq 1$$
This defines a continuous path linking $x$ and $y$ (actually, it is the concatenation of $[x,z]$ and $[z,y]$).
Because $z \notin L_i$ for all $i$, that means that $x_i$ can't belong to the segment $[x,z]$. Because $z \notin L'_i$ for all $i$, that means that $x_i$ can't belong to the segment $[z,y]$. That means that the image of $p_z$ is contained in $X$, and this concludes the proof.
A: Using the argument I'm going to explain, it suffices to show that $\mathbb{R}^3\setminus \{p\}$ is path-connected. Your statement will be proved by induction then. So, let's get to the argument.
As you said, for any two points $x,y \in \mathbb{R}^3$, consider the line segment joining them, i.e. $l(\lambda)=\lambda x + (1-\lambda) y$ for $0 \leq \lambda \leq 1$. Then as you said, if $p \in l(\lambda)$, there exists a unique $0 < \lambda_0 < 1$ such that $l(\lambda_0)=p$. Now take $\lambda_{\star} = \frac{1}{2}\min(\lambda_0,1-\lambda_o) >0$ and define a new continuous path by
$$l_{\text{new}}(t)=\begin{cases}l(t) & t\in[0,1]\setminus(\lambda_0-\lambda_{\star},\lambda_0+\lambda_{\star})\\c(t) & t\in[\lambda_0-\lambda_{\star},\lambda_0+\lambda_{\star}]\end{cases}$$
where $c: [\lambda_0-\lambda_{\star},\lambda_0+\lambda_{\star}] \to \mathbb{R}^3$ is a curve that I'm going to construct now. Take the substitution $s: [\lambda_0-\lambda_{\star},\lambda_0+\lambda_{\star}] \to [0,1]$ given by $s(t) = \frac{1}{2\lambda_{\star}}(x-\lambda_0+\lambda_{\star})$.
Consider $c_0: [\lambda_0-\lambda_{\star},\lambda_0+\lambda_{\star}] \to \mathbb{R}^3$ given by $c_0(t) = \big(\cos(\pi s(t)),\sin(\pi s(t)),0\big)$  which gives the upper half of the unit circle in the $xy$-plane. 
Now let's assume $v=\frac{1}{2}\big(l(\lambda_{0}+\lambda_{\star})-l(\lambda_{0}-\lambda_{\star})\big)\in\mathbb{R}^3$ has the coordinates $v=(a_0,b_0,c_0)$. Take $w=(b_0,-a_0,0)$. Hence, $v \cdot w = 0$ and they're orthogonal. The ordered set $$\{\frac{v}{\|v\|},\frac{w}{\|w\|},\frac{v\times w}{\|v\times w\|}\}$$ is now a right-handed orthogonal frame for $\mathbb{R}^3_p$. Consider the orthogonal transformation (rotation) that sends $\{e_1,e_2,e_3\}$ to $\{\frac{v}{\|v\|},\frac{w}{\|w\|},\frac{v\times w}{\|v\times w\|}\}$ and call it $R$. Now define
$$c(t) = \|v\|Rc_0(t) + l(\gamma(t))$$
where $\gamma: [\lambda_0-\lambda_{\star},\lambda_0+\lambda_{\star}] \to [0,1]$ is a reparametrization such that 
$$l\big(\gamma(\lambda_0-\lambda_{\star})\big)=l(\lambda_0-\lambda_{\star})-v$$
$$l\big(\gamma(\lambda_0+\lambda_{\star})\big)=l(\lambda_0+\lambda_{\star})+v$$
$\gamma$ isn't a linear function, but it can be chosen to piecewise-linear. These conditions guarantee that $l_{\text{new}}$ is continuous.
Geometrically, I'm cutting a small enough portion of the line segment joining $x$ and $y$ at the point $p$ and instead I'm gluing a properly rotated half circle centered at $p$ of radius $\|v\|$ to the endpoints of the segment I removed. $\lambda_{\star}$ has been chosen such that both $\lambda_0+\lambda_{\star}$ and $\lambda_0-\lambda_{\star}$ belong to $[0,1]$.
A: Here's a simpler approach.
Let a,b be any two points in the complement.
Pick any plane passing through ab.
That plane may contain some of the excluded points.
On that plane there are uncountable many arcs of a circle from a to b.
Those arcs are pairwise disjoint except at the endpoints.
As there are only countably many points that are excluded,
it is possible to find an arc from a to b that misses all the excluded points.
Thus there is a path from a to b. 
