Suppose $A$ is a closed subset of $(E,d)$. How to argue that if there is an open ball $B(x,\epsilon)\subseteq A^c$ then $inf \{d(x,y):y\in A \}\geq \epsilon$?
It seems to be very intuitive to me, but I can't prove it. I feel that I have to use triangle inequality, but how? Do I need to create another open ball centered in $y\in A$?
Suppose not, so that $\inf\{d(x,y):y\in A\}=\epsilon'<\epsilon$. By definition of infimum, there exists some $y\in A$ such that $d(x,y)<\epsilon$. By definition of $B(x,\epsilon)$, this means $y\in B(x,\epsilon)$. Then $y\in A\cap B(x,\epsilon)$, but this set is empty by assumption (as $B(x,\epsilon)\subseteq A^c$), a contradiction.