Any subcollection $\mathcal A_1$ of a locally finite collection $\mathcal A$ of subsets of $X$ is locally finite. Any subcollection $\mathcal A_1$ of a locally finite collection $\mathcal A$ of subsets of $X$ is locally finite.
Proof
Let $\mathcal A_1$ be any subcollection of $\mathcal A$ i.e., $\mathcal A_1\subset \mathcal A.$
Let $x\in X$ and $\mathcal A$ is locally finite,  so there exists a nbhd $U$ of $x$ that intersects only finitely many elements $A_1, A_2, A_3,..., A_n$ of $\mathcal A$.
More precisely,$A_i\cap U\neq \phi$,where $1\le i \le n $.
Since,$\mathcal A_1\subset \mathcal A.$,so if any element of {$A_i:1\le i \le n$} is in $\mathcal A_1,$ so $U$ intersects only finitely many elements of $\mathcal A_1$,making it locally finite collection of subsets of $X$.
Please check the proof critically up to here, especially the use of quantifiers.. If there is some scope of improvement in the proof or some elegant method other than this please let me know...
My question is "What if no element of {$A_i:1\le i \le n$} is in $\mathcal A_1$?"
 A: Your proof is okay.
You could also formulate it like this:
Collection $\mathcal A$ is locally finite if and only if for every $x\in X$  a neighborhood $U$ of $x$ exists that satisfies:$$\{A\in\mathcal A\mid A\cap U\neq\varnothing\}\text{ is a finite set}\tag1$$
From $\mathcal A_1\subseteq\mathcal A$ it follows directly that: $$\{A\in\mathcal A_1\mid A\cap U\neq\varnothing\}\subseteq\{A\in\mathcal A\mid A\cap U\neq\varnothing\}\tag2$$so if $\mathcal A$ is indeed locally finite then from $(1)$ and $(2)$ it follows directly that also:$$\{A\in\mathcal A_1\mid A\cap U\neq\varnothing\}\text{ is a finite set}\tag3$$
We conclude that $\mathcal A_1$ is also locally finite.
Concerning your question: 
If $\{A\in\mathcal A_1\mid A\cap U\neq\varnothing\}$ is empty (in your words no element of $\{A_i\mid 1\leq i\leq n\}$ is in $\mathcal A_1$) then the set mentioned under $(3)$ is empty hence is finite.
So nothing is wrong then with the reasoning.
The less elements a collection $\mathcal A$ contains the more likely it is that the collection is locally finite. 
In specific the empty collection is locally finite.
