Is a finite metric space complete? If $(X,d)$ is a metric space with a finite number of elements, is $X$ complete?
I know that every finite metric space is compact, and also compact spaces are complete. So, is this true?
 A: Here is a quick way to see this is true:
Since $X$ is finite, you can find a distance $\rho > 0$ such that no two points are within $\rho$ of each other. For instance $\rho = \frac{1}{2} \min_{x \neq y} d(x,y)$ works. Notice this $\rho$ is strictly greater than $0$ since we're taking the minimum of finitely many nonzero things.
But now fix a cauchy sequence in $X$. Eventually the terms get closer together than any $\epsilon$. In particular, they get closer together than $\rho$. After this point, there is no choice, the sequence must be constant. Then it clearly converges.
The argument for compact spaces is a variation on this theme, so your higher level argument more or less agrees with this one once you unpack the content of the theorem.

I hope this helps ^_^
A: Direct proof.
Let $r=\min\{d(x,y):x,y\in X, x\ne y\}$. Then $r>0$. Suppose $(x_n)$ is a Cauchy sequence. Then there exists $k$ such that, for $m,n\ge k$, $d(x_m,x_n)<r$.
This implies that the sequence is constant from $k$ onwards.
A: It's easy to see that what you actually have here is a discrete space.  Hence every Cauchy sequence is eventually constant.  Hence convergent.
