I came to this question in the course of trying to prove that if $X$ is a paracompact locally Euclidean Hausdorff space with at most countably many components, then it is second countable. I was able to reach the hypothesis in the thread title, then got stuck. I know this problem in general has been addressed before, but I'm wondering if my approach is valid.
The following is what I'd like to clarify, I'm not sure if it's legitimate. Pick any $U_0 \in \mathcal{U}$. There are $n_1$ sets in $\mathcal{U}$ that intersect $U_0$, call them $U_1,...,U_{n_1}$. There are $n_2$ sets that intersect any of the $U_0,...,U_{n_1}$, call them $U_{n_1+1},...,U_{n_1+n_2}$. Proceed this way, letting the sets $U_{n_1+...+n_k+1},...,U_{n_1+...+n_k+n_{k+1}}$ be the $n_{k+1}$ sets that intersect any of the preceding $U_i$.
Now, I argue that the open set $U = \bigcup_{i}U_i$ = X. Suppose it does not. For any $x \in U^c$, there is some $U_x \in \mathcal{U}$ that contains $x$, and it must have failed to appear in the list $U_i$; in fact, by construction it must not intersect any of the $U_i$ so $x \in U_x \subset U^c$. Thus $U^c$ is open, so $X$ is the union of two disjoint open sets ($U$ and $U^c$) and thus disconnected, which is a contradiction.
Local finiteness of $\mathscr{U}$ does not ensure that each member of $\mathscr{U}$ intersects only finitely many other members of $\mathscr{U}$: that's a stronger property.
However, you know that your space is locally separable, so it has a locally finite open cover $\mathscr{U}$ whose members are separable. Let $W$ be any separable open set, and let $D$ be a countable dense subset of $W$. For each $x\in D$ the set $\operatorname{ST}(x,\mathscr{U})=\{U\in\mathscr{U}:x\in U\}$ is finite, and
$$\{U\in\mathscr{U}:U\cap W\ne\varnothing\}=\bigcup_{x\in D}\operatorname{ST}(x,\mathscr{U})\;,$$
so $W$ intersects only countably many members of $\mathscr{U}$. Let
$$\operatorname{st}(W,\mathscr{U})=\bigcup\{U\in\mathscr{U}:U\cap W\ne\varnothing\}\;;$$
$\operatorname{st}(W,\mathscr{U})$ is the union of countably many separable sets, so it is also separable. Now fix $U_0\in\mathscr{U}$; $U_0$ is separable. Given a separable open set $U_n$ for some $n\in\Bbb N$, let $U_{n+1}=\operatorname{st}(U_n,\mathscr{U})$; then $U_{n+1}$ is also a separable open set. Let $W(U_0)=\bigcup_{n\in\Bbb N}U_n$; $W$ is a separable open set.
Show that $\{W(U):U\in\mathscr{U}\}$ is a partition of $X$ into open subsets, and conclude that $X=W(U)$ for each $U\in\mathscr{U}$.
Further deduce that $X$ is the union of countably many members of $\mathscr{U}$.
Finally, note that we could have assumed additionally that each $U\in\mathscr{U}$ was metrizable and hence Lindelöf, and deduce that $X$ is Lindelöf.