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Let $X$ be a Polish space (separable and completely metrizable topological space).

Question: Let $U$ be an open subset in $X$ and let $a,b$ be distinct points in $U.$ Prove that there are open sets $U_a \ni a$ and $U_b \ni b$ such that $\overline{U}_a\subseteq U, \overline{U}_b\subseteq U$ and $\overline{U}_a \cap \overline{U}_b = \emptyset.$

My attempt: Since $U$ is open, there exist $V_a \ni a$ and $V_b \ni$ such that $V_a\subseteq U$ and $V_b \subseteq U.$

Since $a\neq b$, by Hausdorff property of $X,$ there exist open sets $W_a \ni a$ and $W_b \ni b$ such that $W_a \cap W_b =\emptyset.$

Let $U_a = V_a \cap W_a$ and $U_b = V_b\cap W_b.$ Clearly $U_a \ni a$ and $U_b \ni b.$

However, I am not able to show that $\overline{U}_a\subseteq U, \overline{U}_b\subseteq U.$

Is my working correct?

Any hint would be appreciated.

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All you need is that it is metrizable. Take a metric compatible with the topology, and let $U_a$ and $U_b$ be open balls centred at $a$ and $b$ respectively with sufficiently small radius.

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As X is Hausdorff, for distinct a,b,
some disjoint open W,V subset U with a in W, b in V.

As X is regular, there's some open W',V' with
a in W' subset closure W' subset W,
b in V' subset closure V' subset V. W,V are the sets you want.

Thus in regular Hausdorff spaces, distinct points are completely
separated, ie have disjoint closed nhoods.

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