Locally finite open cover of precompact sets in a connected space is countable Let $\mathcal{U}$ be a locally finite open cover of  a connected topological space $X$ such that every $U \in \mathcal{U}$ is precompact and nonempty. We have that every $U$ in $\mathcal{U}$ intersect only finitely many other element of $\mathcal{U}$ (because locally finiteness + precompact sets).
Question: Does $\mathcal{U}$ is a countable open cover?
I think yes, here below an attempt to prove it.
Fix $U_0 \in \mathcal{U}$ and consider $\mathcal{V} \subseteq \mathcal{U}$ defined as the smallest collection of open sets such that:

*

*Contains $U_0$;

*If $V \in \mathcal{V}$ then $\mathcal{V}$ include the finite set $\{U \in \mathcal{U} \, \colon V \cap U \neq \emptyset\}$,

this definition yields a countable subcover of $\mathcal{U}$ (because $X$ is connected).
Are the two covers the same cover? Take $U \in \mathcal{U}$ and $x \in U$, we have that exists $V \in \mathcal{V}$ such that $x \in V$ so $U \cap V \neq \emptyset$ which implies that $U \in \mathcal{V}$. So yes, the two covers are the same cover.
 A: Take $\mathcal{V}_0 = \{U_0\}$ and for every $n \ge 1$ consider
$$\mathcal{V}_{n+1} =\{U \in \mathcal{U} \mid \exists V \in \cup_{\,i=0}^{\,n} \mathcal{V}_i \;\;\text{s.t.} \; U \cap V \neq \emptyset \} - \cup_{\,i=0}^{\,n} \mathcal{V}_i.$$
We see that $\mathcal{V}_i$ is finite for all $i \ge 0$.
If $\mathcal{V}_0, \dots, \mathcal{V}_n$ are finite sets then $\cup_{\,i=0}^{\,n} \mathcal{V}_i$ is also finite.
Using the fact that every $V \in \cup_{\,i=0}^{\,n} \mathcal{V}_i \subseteq \mathcal{U}$ intersect only finitely many other element of $\mathcal{U}$ then also $\mathcal{V}_{n+1}$ is finite and follows by induction that $\mathcal{V}_i$ is finite for all $i \ge 0$.
We have that $\bigcup_{\,i=0}^{\,\infty} \mathcal{V}_i$ is a countable cover of $X$, it is a cover because $X$ is connected.
If we prove that $\mathcal{V} = \bigcup_{\,i=0}^{\,\infty} \mathcal{V}_i$ then we are done (see definition of $\mathcal{V}$ in the question). The non trivial inclusion is "$\subseteq$". Take $V \in \mathcal{V}$ with $V \neq U_0$ and consider $x \in V$ and $y \in U_0$. The space $X$ is connected so exists a finite sequence $V_0, \dots, V_k \in \mathcal{V}$ such that $V_i \cap V_{i+1} \neq \emptyset$ for all $i \in \{0,\dots,k-1\}$ and $x \in V_0$, $y \in V_k$, this proves that $V$ is an element of $\bigcup_{\,i=0}^{\,\infty} \mathcal{V}_i$.
