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Is it always possible to find a countable open locally finite refinement of a given open cover in a paracompact second-countable space? I guess, it is. But I don't know how to prove it.

Edit: Arthur Fischer's answer gave me the idea to a much easier proof: We just show that every open locally finite cover of the space is already countable. So let $(U_i)_{i\in I}$ be an open locally finite cover of $X$ and $D$ countable with $\overline{D}=X$. Without loss of generality $U_i\ne \emptyset$ for all $i\in I$. Let $i\in I$ then $D\cap U_i\ne\emptyset$ because otherwise $\overline{D}\cap U_i=\emptyset$, a contradiction to $D$ being dense. For each $d\in D$ there are only finitely many $i\in I$ with $d\in U_i$ because of $(U_i)_{i\in I}$ being locally finite. Let $I_d$ be the set of those $i$s. Since $D\cap U_i\ne\emptyset$ we have that each $i\in I$ is in some $I_d$. Hence $I=\bigcup_{d\in D} I_d$ is countable.

How does that solve my question? Well, given an open cover in a paracompact second-countable space implies because of paracompactness that there is a open locally finite refinement. By what I've shown above this refinement has to be countable.

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Yes, because every second-countable (in fact every separable) paracompact space is Lindelöf.

Suppose that $X$ is a separable paracompact space with countable dense set $D$. If $\mathcal{U} = \{ U_i \}_{i \in I}$ is an open cover, by paracompactness there is a locally finite refinement $\{ V_i \}_{i \in I}$ with property that $\overline{V_i} \subseteq U_i$ for all $i \in I$. Consider $I_0 = \{ i \in I : V_i \cap D \neq \emptyset \}$. Clearly $I_0$ is countable (since each element of $D$ is contained in finitely many $V_i$). Now note that $$X = \overline{D} = \overline{\bigcup_{i \in I_0} ( D \cap V_i )} = \bigcup_{i \in I_0} \overline{ D \cap V_i} = \bigcup_{i \in I_0} \overline{ V_i } \subseteq \bigcup_{i \in I_0} U_i.$$ (The equality $\overline{\bigcup_{i \in I_0} ( D \cap V_i )} = \bigcup_{i \in I_0} \overline{ D \cap V_i}$ holds because $\{ D \cap V_i : i \in I_0 \}$ is locally finite.) Therefore $\{ U_i \}_{i \in I_0}$ is a countable subcover of $\mathcal{U}$.

So given an open cover $\mathcal{U}$ of $X$, find a locally finite refinement $\mathcal{V}$ of $\mathcal{U}$, and then a countable subcover $\mathcal{W}$ of $\mathcal{V}$.

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