Definition of Open set. I'm confused about the definition of Open set
"A set Z is open, if for every element z in Z there's a r>0 such that there's an open ball contained with radius r in Z" 
I read the proof about "An open ball is an open set" 
What I don't understand is: "what if we take a radius so big such that the open ball isn't completely contained in Z" 
I just want to understand what I'm getting wrong.
 A: ... then you take a smaller radius.  
A set $Z$ is open if for every point $x$ you can find an open ball (which might be very  small) centered at $x$ that is contained in $Z$.  
It doesn't mean every ball will work.  It only means some balls will work.  Actually it only needs to mean at least one ball will work.  [That usually means the smaller balls will work too but sometimes there aren't any smaller balls (don't worry about that now.)]
If for example if you take the interior of a unit circle in a plane centered at $(0,0)$.  That's open because "it has fuzzy edges"...  Anyway if you take any point in it.  Say $(0, 0.9999)$ then I can find an open ball centered at $(0,0.9999)$ entirely in the circle.  To do that I have to take a radius $r \le 0.0001$; but I can do it if I take a radius small enough.
Now what if instead I had taken a radius of $27$ and that's way too big to fit in the circle?  Well, I shrug my shoulders and give the person asking a confused look and say "Who cares about that radius? I can take a radius of $0.0001$ and that does work.  I dont care about radii that don't work.  I just care that there are radii the do work."
A: By definition, a set $U$ in a metrix space $(X,d)$ is open if for every $x \in U$, there exists r>0 such that $B(x,r) \subseteq U$
In your case, in order to prove $B(x,r)$ is open, for every $y \in B(x,r)$, we choose a suitable $s>0$ for which $B(y,s) \subseteq B(x,r)$.    

what if we take a radius $s$ so big such that the open ball isn't completely contained in the set ?

Why we need such a big $s$ ? By definition, we want there is such an s , so your case , $$s \leq r-d(x,y)$$ works!
A: It seems that your specific question has been answered. It seems to me to be more important to point out that your version of the definition is wrong! If you want to understand a definition first  you have to get the definition straight... (It may well be that what you meant by the definition is correct, but if so you didn't say what you meant.)
This is important -  if you want to understand these things you need to be much more careful about the language.  You say this:


Def 1. A set Z is open, if for every element z in Z there's a r>0 such that there's an open ball contained with radius r in Z.


Fixing up the English a little,


Def 1'. A set Z is open, if for every element z in Z there's a r>0 such that there's an open ball  with radius r contained in Z.


This should look suspicious, because you say "for every element z in Z" but then you never mention z again in the rest of the definition.
The correct definition, probably (I hope) what you meant, is this:


Def 2. A set Z is open, if for every element z in Z there's a r>0 such that the  open ball with radius r and center z is  contained in Z.


Let's say the open ball with center $x$ and radius $r$  is $B(x,r)$, to make things easier to state. Then Def 2 is the same as

Def 2'. A set $Z$ is open if for every $z\in Z$ there exists $r>0$ such that $B(z,r)\subset Z$. 

An example showing that the two definitions are not the same: Let's talk about suubsets of $\Bbb R$ with the standard metric $d(x,y)=|x-y|$. Define $$Z=(-1,1)\cup\{2\}=\{z\in\Bbb R:|z|<1\text{ or }z=2\}.$$Then $Z$ is not open, according to the correct definition, because if we say $z=2$ then $z\in Z$ but there does not exist $r>0$ with $B(z,r)\subset Z$.
But $Z$ does satisfy Def 1. First, $B(0,1)=(-1,1)\subset Z$. So the statement 
(i) "there's a r>0 such that there's an open ball contained with radius r in Z"
is true (proof: let $r=1$).
And since (i) is true, it follows that
(ii) "for every element z in Z there's a r>0 such that there's an open ball contained with radius r in Z"
is true! (Because statement (i) simply doesn't mention $z$, the fact that (i) is true implies that (i) is true for every $z\in Z$.)
So Def 1 says that $Z$ is open, while Def 2 says it's not open.
If you followed that then at this point you're saying that's not what you meant. Fine, but that's exactly the important point: You need to be much more careful about the language or you have no chance with this advanced math stuff.
Note: It seems possible that English is not your native language.  If so that's a valid excuse for giving Def 1 when it really makes no sense, should be Def 1'. But if you think that's an excuse for the whole thing you're fooling yourself! The difference between Def 1' and Def 2 or Def 2' is not just a matter of poor English expression, it's a matter of logic. Saying "for every z in Z" and then  never mentioning z again makes no sense in any language.
