Every locally finite family of non-empty subsets of a Lindelöf space is countable. 
I just don't understand the conclusion of the lemma: $|\mathcal{A}| \le \aleph_0$. I think it's related with the fact that every member of $\mathcal{U}$ meets only finitely many members of $\mathcal{A}$, while every member of $\mathcal{A}$ meets at least one member of $\mathcal{U}$ (could be a countable number of memebrs), but it's not true in general that if $S$ is a set and $A, B$ are families of subsets of $S$ such that every member of $A$ intersects at least one member of $B$ and every member of $B$ intersects only a finite number of members of $A$, then $|A| \le |B|$ (for example, take $S = \{1, 2\}$ and $A = \{\{1\}, \{2\}\}, B = \{\{1, 2\}\}$.)
 A: Yes, it is related to the fact that every member of $\mathcal{U}$ meets only finitely many members of $\mathcal{A}$ - although for the conclusion it would suffice that every $U\in \mathcal{U}$ meets only countably many members of $\mathcal{A}$.
Since every member of $\mathcal{A}$ meets some $U\in \mathcal{U}$, we can choose one such $U$ for every $A\in \mathcal{A}$. That gives us a map $\tau \colon \mathcal{A} \to \mathcal{U}$. Now we can write
$$\mathcal{A} = \bigcup_{U\in \mathcal{U}} \tau^{-1}(\{U\}),$$
which gives us $\mathcal{A}$ as a union of countably many finite sets. That shows $\mathcal{A}$ is countable, since countable unions of countable sets (in particular countable unions of finite sets) are countable.
Concerning your following remark, indeed, for finite sets, an analogous inequality does not follow, it's the arithmetic of infinite cardinals, where $\kappa + \lambda = \kappa\cdot\lambda = \max \{\kappa,\lambda\}$, so a union of $\kappa$ many sets of cardinality $\leqslant \kappa$ has cardinality $\leqslant \kappa$.
