Proof: Open Ball is open I have a question to the following proof: 
Theorem: For $(X,d)$ a metric space, $a \in X$, $r\geq 0$, then the open ball $U_r(a)$ is a open set. 
Proof: For $r=0$ is $U_0(a) =\emptyset$ and per defintion is the empty set open. Let $r> 0$, $x \in U_r(a)$ and $\delta:= r-d(x,a) >0$.Than we need to proof that $U_\delta \subseteq  U_r(a)$ follows. 
Let $y \in U_\delta(x)$ than the Triangle inequality says
\begin{align*}
d(y,a) \leq d(y,x) + d(x,a) < \delta + d(x,a) = r -d(x,a) + d(x,a) 
\end{align*}
this means $d(y,a) < r$ and therefore $U_\delta \subseteq  U_r(a)$. Then $U_r(a)$ is open. $\square$
My question is why can we choose $\delta$? Is it because we know that between two real numbers always is another real number. Or what is the theorem that stands behind that?
 A: Since your choice of $\delta$ is $\delta=r-d(x,a)$, then I'd say that we can choose $\delta$ this way because, given two real numbers $x$ and $y$ such that $x>y$, then there is a real number $z>0$ such that $x-y=z$.
A: If your question is "why can we choose a $\delta$?"
The definition of an open set $A$ is as follows: for every $a \in A$, there exists a $\delta > 0$ such that $U_\delta(a) \subset A$. 
The definition says a $\delta$ exists, so you only have to take 1, so you can choose it.
If your question is "why can we choose that specific $\delta$?"
By definition $d(x, a) < r$, so $r - d(x, a) > 0$ and that is the only condition on $\delta$.
A: Let $ A $ be an open ball in the metric space $ (X,d) $. The, from definition of open ball, we must have $$ A=B_{\varepsilon} (x)=\{ y \in X: d(y,x)<\varepsilon \}$$
Let $ w \in A \iff d(w,x)<\varepsilon \Rightarrow 0<\varepsilon -d(w,x) $
Hence, we choose $ \delta=-d(w,x)+\varepsilon >0 $ and we construct a ball centered at each member of $ A $ which is $ w $ as follows $$B_{\delta} (w)=\{ y \in X: d(y,w)<\delta \}$$
Let $z \in B_{\delta} (w)  $. Then $ d(z,w)<\delta=-d(w,x)+\varepsilon \Rightarrow d(z,w)< -d(w,x)+\varepsilon \Rightarrow d(z,w)+d(w,x)<\varepsilon $
From the third axiom of definition of metric space we must have $$d(z,x)\le d(z,w)+d(w,x)<\varepsilon\Rightarrow d(z,x)<\varepsilon \Rightarrow z\in A \Rightarrow B_{\delta} (w) \subset B_{\varepsilon} (w)=A$$
Hence, $ A $ is an open set.
