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?

First, your answer is incorrect. $S$ could also have $0$ points.
Let's prove that $S$ cannot have more than 1 point.
Suppose that $S$ contains $n \geq 2$ distinct points. Write $S = \{x_1, x_2, ..., x_n\}$.
Let $\kappa = \min\limits_{2 \leq i \leq n} d(x_1, x_i)$. Then $\kappa > 0$. Let $\epsilon = \frac{\kappa}{2}$.
Let $A = B_\epsilon(x_1)$, and let $B = \bigcup\limits_{i = 2}^n B_\epsilon(x_i)$.
Then $A$ and $B$ are clearly open sets. We have $x_1 \in A$ and $x_2, ..., x_n \in B$, so $A$ and $B$ are nonempty and their union contains $S$.
Finally, $A$ and $B$ are disjoint. For if we have some $z \in A \cap B$, then $d(x_1, z) < \epsilon$ and $d(z, x_i) < \epsilon$ for some $i \geq 2$. Then we have $d(x_1, x_i) \leq d(x_1, z) + d(z, x_i) < \epsilon + \epsilon = \kappa \leq d(x_1, x_i)$, which is a contradiction.
So $S$ is not connected. Since every path-connected set is connected, $S$ is not connected.
Note that there is no need to restrict $S$ to being a subset of $\mathbb{R}^n$. $S$ could be any metric space (or even any Hausdorff space, though the Hausdorff space argument is a slight modification of the above).
