Open sets Are balls? Is any open set a ball? 
So instead of open sets we can talk about balls or disks?
Is that correct? Because ihave seen definitions with open sets Like the definition of a surface.So is it sufficient instead of talking about  generally open sets simplify things thinking it as b
So the question is: given a subset $U \subseteq \mathbb{R}^n$, we say that a subset $K$ of $U$ is open in $U$ if there exists an open set $M$ of $\mathbb{R}^n$  such that $K = M \cap U$ .  In the definition, can we change "open set $M$ of $\mathbb{R}^n$" to "open ball $M$ in $\mathbb{R}^n$?"
 A: No. Draw some simple figures in the plane as such: 
But by the definition of an open set, to each point $x \in U$, there exists an open ball around that point completely contained in the set.
Thus,
$$U = \bigcup_{x \in U} B_{\epsilon(x)}(x)$$
where $\epsilon(x)$ is a radius that keeps the open ball around $x$ contained in $U$.
Edit: Your edit is a question about the relative topology. Take an open subset $K$ of $U$ above that doesn't look like a ball and notice that no ball $M$ exists such that $K = U \cap M$. Do some reading on the concept of the relative topology. Thus, no it does not suffice.
A: There are many definitions of open sets. If you have a topology, then all elements of it are defined to be open, and they do not need to be balls.
If you have a metric, then a set $A$ is open if for all $x\in A$, there exists an open ball $B(a,r)$ (an open ball is defined as $B(a,r)=\{x: d(x, a)<r\}$) such that $x\in B(a,r)\subseteq A$.
This means that in a metric space, all open balls are also open sets, but not all open sets are open balls. For example, $(0,1)\cup (3,4)$ is not an open ball, but it is an open set.
A: Take $U$ to be the unit circle $\{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \}$.  Let $K$ be $U$ with the point $(0,1)$ deleted.  Then (i): $K$ is open in $U$, but (ii): you can't get $K$ by intersecting any open disc in the plane with $U$.  You can see what I'm talking about by drawing a picture.  
A: Generally, in an arbitrary metric space $(X,d) $ is not true that every open is open ball. Certainly it is true that the open balls are open in the topology induced by the metric, but if $(X,d)$ is the discrete metric space, that is, $d (x,y)=1 $ if $x \neq y$ and $d(x,y)=0$ if $x=y$, then $B(x,r)=\lbrace x \rbrace$ if $0 < r \leq 1$ and $B(x,r)=X$ if $r > 1$, in particular $\lbrace x \rbrace \in \mathcal{T}_d$ for every $x \in X$, where $\mathcal{T}_d$ is the topology induced by metric, therefore $P(X)=\mathcal{T}_d$, and if $|X|>2$ and $x \neq y$ we have that $\lbrace x,y \rbrace$ is open but not an open ball.
