Cauchy sequence in finite space is eventually constant Let $(F,d)$ be a metric space, with $F$ a finite set, and let $(a_n)$ be a Cauchy sequence in $F$.
Then $(a_n)$ is eventually constant.
Definition: We say that $(a_n)$ is eventually constant if $\exists N \in \mathbb{N} \ni \forall n \in \mathbb{N},[n \ge N \Rightarrow a_n=a_N]$.
Lemma: If $(X,d)$ is a metric space and the points $x,y \in X$ satisfy $\forall \varepsilon \in (0,\infty), [d(x,y)<\varepsilon]$, then $x=y$.
Suppose $x \ne y$, then by definition of a metric space, it follows that $d(x,y) \ne 0$. Because $\forall x,y \in X$ we have $d(x,y) \ge 0$, let $\varepsilon=\frac{d(x,y)}{2}>0$. Thus $\exists \varepsilon \in (0,\infty)$ for which $d(x,y)>\varepsilon$. Therefore, $d(x,y)<\varepsilon \Rightarrow x=y$.
Proof. Suppose $(a_n)$ is a Cauchy sequence in $F$. Then for any $\varepsilon >0$, there exists $N \in \mathbb{N}$ such that for all $n,m \in \mathbb{N}$ we have $d(a_n,a_m)<\varepsilon$ whenever $(n \ge N) \wedge (m \ge N)$. By the lemma, this implies that $a_n=a_m$ for all $n,m \ge N$. And by definition this means that $a_n=a_m=a_N$ for all $n,m \ge N$, thereby showing that $(a_n)$ is eventually constant.
Edit
Define $\delta_F=min\{d(x,y)|x,y \in F, x \ne y\}$. As $F$ is finite, then $\delta_f$ is well-defined. By properties of a metric space, if $x \ne y$, then $d(x,y) \ne 0$. And since $\forall x,y \in F, d(x,y) \ge 0$, it follows that $\delta_F >0$.
Because $(a_n)$ is Cauchy in $F$, then $\exists N \in \mathbb{N}$ such that $\forall n,m \in \mathbb{N}$, we have $d(a_n,a_m)<\delta_F$ whenever $(n \ge N) \wedge (m \ge N)$. But by definition of $\delta_F$, if $d(a_n,a_m)<\delta_F$ for all $n,m \ge N$, then $a_n=a_m$. Therefore, $(a_n)$ is eventually constant.
 A: You need a different lemma that actually uses the finiteness of the space $F$.
For a finite metric space $F$ it holds that:
$$\exists \delta_F >0: \forall x,y \in F: (d(x,y) < \delta_F \to x=y)\tag{1}$$
Proof: we can define $\delta_F=\min\{d(x,y)\mid x, y \in F, x \neq y\}$, which is well-defined as $F$ is finite, so we take a minimum of finitely many reals, and $\delta_F>0$ as $d$ is a metric (as opposed to a pseudometric e.g.); it clearly works for $(1)$.
Now apply the definition of Cauchyness to $\varepsilon$ equal to this $\delta_F$ and finish as you do.
A: No. You forgot that $N$ depends on $\varepsilon$ and can grow arbitrarily large with $\varepsilon$ going to $0$.
Rather argue like this. Set
$$
\Gamma := \min \lbrace d(x, y) : x, y \in F, ~ x \neq y \rbrace > 0.
$$
The minimum exists, because $F$ is finite. Then, there exists $N \in \mathbb{N}$ such that
$$
d(x_n, x_m) < \Gamma
$$
id $n, m \geq N$. But if $d(x_n, x_m) < \Gamma$, then $x_n = x_m$ for $n \geq N$ according to the definition of $\Gamma$. This concludes the proof.

The key is that is $F$ is finite which proves that $\Gamma$ is well-defined.
