$X=\bigcup_{i}A_i$ is a "basic" cover if for all $U\in X$, $U$ is open in $X$ iff $\forall i,U\cap A_i$ is open in $A_i$ $X=\bigcup_{i}A_i$ is a "basic" cover if for all $U\in X$, $U$ is open in $X$ iff $\forall i,U\cap A_i$ is open in $A_i$. This is a definition and the only reason it is the one to appear in the subject is that the definition seems to not exist, partially because the term "basic" rings no bell. 
The question asked then is:

Prove that $X=\bigcup_{i}A_i$ is a basic cover iff for all $U\subset X$, if $U\cap A_i$ is open in $A_i$ for all $i$ then $U$ is open.

I cant seem to have any approach. I am not even sure I have percepted the definition right. Doesn't $U\in X$ simply mean $U=A_i$ for some $i$? Does $U\in X$ mean a union? I am perplexed and would appreciate it if you could assist me.   
 A: First, let me give you some details about the definition:

Definition. We say that $X=\bigcup\limits_{i \in I}A_i$ is a basic cover of $X$ if for all $U \subset X$, 
  $$U \text{ is open in } X
\;\;\iff\;\;
\forall i \in I,\; U\cap A_i \text{ is open in } A_i.$$ 

You have to write $U \subset X$ and not $U \in X$. Here $U$ is a subset of $X$, not an element.
Here is a (non-)example. For instance, $\Bbb R = \bigcup_{i \in \Bbb R} \{i\}$ is not a basic cover since $U=\{5\}$ is not open in $\Bbb R$, but
$U \cap \{i\}$ is always open in $\{i\}$.
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Let's turn to the given problem.

Prove that $X=\bigcup_{i}A_i$ is a basic cover iff for all $U\subset X$, if $U\cap A_i$ is open in $A_i$ for all $i$ then $U$ is open.

Therefore, we have to prove that  $X=\bigcup_{i}A_i$ is a basic cover if and only if we have the following property (let me call it the property (Q)):

For every subset $U \subset X$,
  $$\forall i \in I,\; U\cap A_i \text{ is open in } A_i
\;\;\implies\;\;
U \text{ is open in } X$$

Hence, we have to show that the following property (P) (which is exactly the definiton of "basic cover" given above):

For all $U \subset X$, 
  $$
\forall i \in I,\; U\cap A_i \text{ is open in } A_i
\;\;\iff\;\;
U \text{ is open in } X.$$

is equivalent to the above property (Q).
You can see that (P) and (Q) are very similar:


*

*(P) has the form $\phi \iff \psi$

*(Q) has the form $\phi \implies \psi$


(where $\phi$ denotes "$\forall i \in I,\; U\cap A_i \text{ is open in } A_i$" and $\psi$ denotes "$U \text{ is open in } X$").
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What does it mean to be "open in $A_i$" ?
Let $B \subset X$ be any subset of $X$.

Definition
  A subset $V \subset B$ is said to be open in $B$ iff
  there is an open set $W \subset X$ such that
  $V = W \cap B$.

By the above definition of the subspace topology, we have
$$U \text{ is open in } X \implies \forall B \subset X,\; U\cap B \text{ is open in } B$$
because you just have to take $W=U$ (and here $V=U \cap B$).
Applying this to any $B = A_i$, we get
$$U \text{ is open in } X \implies \forall i \in I,\; U\cap A_i \text{ is open in } A_i$$
which is $\psi \implies \phi$.
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From all this, I think you should be able to prove that


*

*(P) implies (Q) (easy)

*(Q) implies (P) : assume that (Q) is true, i.e. $\phi \implies \psi$ is true. You want to show that (P) is true, i.e. $\phi \iff \psi$ is true.  By what I've said above, $\psi \implies \phi$ is always true. By assumption, you know that $\phi \implies \psi$ is also true. Therefore, you can easily conclude that $$\phi \iff \psi$$ (which is (P)) is also true, as desired.


Therefore (Q) and (P) are equivalent, as desired.
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From very similar arguments, you can replace the open sets $U$ by closed sets $F$ in the definiton, since 
$F$ closed in $X \implies F \cap B$ closed in $B$ for any $B \subset X$. 
