$X$ is a complete metric space. Suppose $K$ is a closed and bounded subset with the property that it contains a countably infinite set of points $\{x_k\}$ with the property that $$d(x_m,x_n) \geq 2\alpha >0$$ for all $n\not =m$. Show that $K$ is not compact.
Proof: Let $\epsilon = \alpha>0$ and $B_\epsilon(x_k)$ be open balls of center $x_k$ with radius $\alpha$. Since each $B_\epsilon(x_k)$ contains only one point $x_k$ in $K$, there are countably infinite such open balls. $K=\bigcup_{i=1}^\infty B_\epsilon(x_k)$, so $K$ is contained in the union of a collection of open subsets of $X$, but $K$ is not contained in the union of finite number of these open subsets.
Is this proof correct?
This is not correct. If you want to show that $K$ is not totally bounded, you need to show that (for some $\epsilon>0$) there does not exist any finite collection of elements $a_1,\dots,a_n\in K$ such that $K$ is covered by the balls $B_\epsilon(a_i)$. So to prove this, you need to start with an arbitrary finite collection of elements $a_1,\dots,a_n\in K$ and prove that the balls $B_\epsilon(a_i)$ cannot cover $K$. You have done no such thing.
(Also, there is no reason for the equation $K=\bigcup_{i=1}^\infty B_\alpha(x_k)$ to be true. But even if it were true, it would not (directly) prove that $K$ is not totally bounded, as you claim.)