locally finite closed coverings are fundamental First of all the definitions, since i'm not english and I don't know if it's clear what I'm talking about. Let's take a topological space $X$
A covering is locally finite if for every $x \in X$, there is a neighbourhood $I_x$ of $x$ that intersects only a finite amount of sets of the covering. The situation I am proposing is about a locally finite covering whose elements are closed sets.
A covering is called fundamental if for every set $\Omega$ in $X$, we have that $\Omega$ is open iff $\Omega \cap A_i$ is open in $A_i$ for every $i\in I$, where $\{A_i \}_{i\in I}$ is the fundamental covering.
I think this theorem is quite simple, because it was proven in a first course of topology. To be precise, it was first proven that every finite closed covering (that is a finite covering composed of closed sets) is fundamental, and then this result was used to prove this stronger version (every finite closed covering is locally finite).
The question is: how can I use the fact that every finite closed covering is fundamental to prove that every locally finite closed covering is fundamental as well?
The problem is that I don't get the proof, cause when he says like "this is open" and "this is closed" I can't understand where he is working, and how I can deduce that if a set is (for example) open in a subset, then it is open in the whole topological space. I know how to conclude that if the subspace is open itself, but I can't see how to apply that here
 A: A cover $A_i, i \in I$ of $X$ is also fundamental iff

A subset $A$ of $X$ is closed iff $A \cap A_i$ is closed in $A_i$ for each $i$.

The proof is obvious.
Also: 

A locally finite family of closed sets $B_i, i \in I ,B_i \subseteq X$ has a closed union $B = \cup_i B_i$ (note that this is an axiom for finite unions).

For suppose $p \in \overline{B}, p \notin B$. As the family is locally finite, there is an neighbourhood $U$ of $p$ such that $I_0 = \{i: U \cap B_i \neq \emptyset \}$ is finite. But then $B_0 = \cup \{B_i: i \in I_0\}$ is closed, as a finite union of closed set, and then $U \setminus B_0$ is a neighbourhood of $p$ that misses all $B_i$, contradicting that $p \in \overline{B}$. So indeed $B$ is closed.
Now to see locally finite closed covers are fundamental: take $A \subseteq X$.
If $B$ is closed, all $A \cap B_i$ are closed in $B_i$ by definition of the subspace topology. If however all $B_i \cap A$ are closed in $B_i$ all sets $A_i = B_i \cap A$ are closed in $X$ by a standard fact, and then $A_i, i \in I$ is a locally finite family (as the larger $B_i$ already is locally finite) and its union $\bigcup_i A_i = \bigcup_i (A \cap B_i) = A \cap (\bigcup B_i) = A \cap X= A$ is closed by the above fact. This shows $B_i, i \in I$ is fundamental.
