Prove compact subsets are not infinite 
If there exists $i>0$ such that for every $x,y\in M$ with $x\neq y$ we have $i\le d(x,y)$, then compact subsets of this metric space are not infinite.

My try: It is known that a discrete space is compact if and only if it is finite. So I try to prove that this metric space is also a discrete space and I got stuck.
 A: (The following answer is a recap of the discussion below the question.)
As you say in the post, it suffices to prove that $M$ is discrete. Evidently your definition of discreteness (for a metric space) is:
Def. $M$ is discrete if any convergent sequence in $M$ is eventually constant (i.e., if $(x_n)$ converges to $x$ then there is some $N$ such that $x_n=x$ for all $n\geq N$).
We can prove that $M$ satisfies this property. Suppose $(x_n)$ converges to $x$. Since $M$ is a metric space, this means that for all $\epsilon>0$, there is some $N$ such that $d(x_n,x)<\epsilon$ for all $n\geq N$. Apply this with $\epsilon=i$. Then we get $N$ such that $d(x_n,x)<i$ for all $n\geq N$. By the assumption on $i$, this means $x_n=x$ for all $n\geq N$, which is what we wanted to show.
Remark. (elaborating on the answer by Paul Frost) Another definition of discreteness, which works for any topological space, is that $M$ is discrete iff for any $x\in M$, the singleton set $\{x\}$ is open. This is also easy to prove from your assumption on $M$. Indeed, if $x\in M$ then $\{x\}$ is the open ball of radius $i$ centered at $x$ (since, for any $y\in M$, $d(x,y)<i$ iff $x=y$).
A: Hint:
Let's try a proof by contradiction. Suppose that $C\subset M$ is infinite and compact in $M$. Since $C$ is infinite, we can construct a sequence $(x_n)_{n\in\mathbb{N}}$ of distinct elements in $C$, right? But since $C$ is compact, that means $(x_n)_{n\in\mathbb{N}}$  has a subsequence $(x_{k_n})_{n\in\mathbb{N}}$ which converges, say $x_{k_n}\to x\in C$. Can we arrive at a contradiction using the notion of convergence and the property in the problem above?
A: Under your assumption $M$ has the discrete topology which means that all one-point subsets $\{x\}$ are open. In fact, in a metric space all open balls $B_r(x) = \{ y \in M \mid d(x,y) < r\}$ with $x \in M$ and $r > 0$ are open subsets. In your case we have $B_i(x) = \{x\}$.
Now let $K \subset M$ be compact. But $\mathcal U = \{ \{x\} \mid x \in K\}$ is an open cover of $K$ and must have a finite subcover. This means that $K$ is finite.
