Equivalent Definitions of a Compact Subset of a Metric Space Let $X$ be a metric space and let $K \subseteq X$. Please consider the following three statements.

  
*
  
*For every collection $\{V_\alpha : \alpha \in \mathcal{A}\}$ of subsets of $X$
  which are both open in $X$ and cover $K$,
  there exists a finite subset $A$ of $\mathcal{A}$ such that $\{V_\alpha : \alpha\in A\}$ cover $K$.
  
*For every collection $\{V_\alpha : \alpha \in \mathcal{A}\}$ of subsets of $K$
  which are both open in $X$ and cover $K$,
  there exists a finite subset $A$ of $\mathcal{A}$ such that $\{V_\alpha : \alpha\in A\}$ cover $K$.
  
*For every collection $\{V_\alpha : \alpha \in \mathcal{A}\}$ of subsets of $K$
  which are both open in $K$ and cover $K$,
  there exists a finite subset $A$ of $\mathcal{A}$ such that $\{V_\alpha : \alpha\in A\}$ cover $K$.

I've convinced myself that 1. and 3. are equivalent. It's also fairly automatic that 2. is implied by either 3. or 1., but I cannot seem to prove that 2. implies either 1. or 3.... Is statement 2., simply, strictly weaker that statements 1. and 3.? Or is there some trick that I'm missing?
Thank you.

This is false. Take $X=\mathbb R$ and $K=\mathbb N$. There is no non-empty open set in $X$ contained in $K$  so 2) is vacuously true. But 1) is false. 
