Proof of compactness by Borel-Lebesgue theorem

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Given the following set

Let $ x \in B[(0,0),1]$ in $\mathbb{R^2} $

(B stands for the closed ball of centre (0,0) and radius 1)

How do I proof that the set $A = B[(0,0),1] - \{x\}$ is not compact by using Lebesgue-Borel definition, i.e. that it doesn't exist a finite subcover for a cover of A?

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For all r > 0, $U_r$ = { a : r < d(x,a) } is open.
Cover the space with { $U_r$ : r > 0 }.

Why not just prove it is not compact since it
is not a closed subset of a Hausdorff space?