Suppose $S \subset \mathbb{R}^n$ is a finite set and is path connected. How many points does $S$ contain? Suppose $S \subset \mathbb{R}^n$ is a finite set and is path connected. How many points does $S$ contain?
My answer: $S$ must contain exactly one point.
Proof: Assume that $S \subset \mathbb{R}^n$ is a finite, path-connected subset with more than one element. By definition of path-connected, there exists a continuous function $f: [0,1] \longrightarrow S$ such that $f(0)=s_1$ and $f(1)=s_2$ for any $s_1,s_2 \in S$. In order for $f$ to be continuous, there then must be uncountably infinite $s_i \in S$ with $f(x_i)=s_i$ for all $x_i$ with $0 < x_i < 1$. This contradicts our assumption that $S$ is finite, so therefore $S$ has exactly one point. $\Box$
Is this an accurate way of stating the reasoning behind this answer? Is there a simpler way of explaining why this is true?

It may be that the infinitely many points between $0$ and $1$ map to only finitely many points on $S$, so more justification is needed. Instead I would say that if $S$ is path connected, it must also be connected. But if $S$ contains more than one point, it is possible to separate $S$ into two open sets, one containing one point and one containing all the others. This is the same as Zanzag's suggestion.
