Prove that discrete metric space is complete

If $x_n$ is a cauchy sequence then, for every $\epsilon>0$ exists an $N \in \mathbb{N}$ such that if $n,m$ are greater than $N$ you have $d(x_n,x_m)<\epsilon$. Now take $\epsilon=1/2$ then, it exists an $N$ such that $d(x_n,x_m)<1/2$ because d is the discrete metric, this is only possible if $x_n$ is a constant for $n$ greater than $N$

Essentially, yes. There is merely the small technicality that a sequence can converge in a discrete metric space if it is just cofinitely constant (i.e. only finitely many terms differ from some specified value).

Yes. If you take $\epsilon <1$, then you will find every Cauchy sequence is a finally constant sequence.