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The quotient map $q:S^2\to X$ is continuous and surjective by the quotient topology, so $X$ is compact as a continuous image of a compact space. To prove the Hausdorff property explicitly, let $[u]\ne [v]$ and set $d=\min (\Vert u-v\Vert ,\Vert u+v\Vert )>0$. Choose $r<d/3$. The saturated open setsare disjoint by the triangle inequality. Each contains both antipodes of every one of its points, so $q^{-1}(qU)=U$ and similarly for $V$. Consequently $qU,qV$ are disjoint open quotient neighborhoods separating the classes. Thus the quotient is compact and Hausdorff. It is the real projective plane; equivalently the metric on the antipodal sphere quotient is $d_X([u],[v])=\min (\Vert u-v\Vert ,\Vert u+v\Vert )$ and induces this topology.