Source: Kechris, pg 26, Theorem 4.25
If $X$ is completely metrizable, so is $K(X)$.
Fix a complete compatible metric $d \leq 1$ on $X$. Let $(K_n)$ be Cauchy in $(K(X),d_H)$, where WLOG we can assume $K_n \neq \emptyset$. Let $K=\overline{T \lim_n} K_n$. We will show that $K \in K(X)$ and $d_H(K_n,K) \rightarrow 0$. Note first that $K= \bigcap_n (\overline{\bigcup_{i=n}^{\infty}K_i})$ and that $K$ is closed and nonempty.
My question: Why is $K$ nonempty?