Every $\epsilon$-neighborhood of a point in a closed ball contains a second point of the ball Let $E$ be a closed ball with centre at $0$ and radius $1$ at $\mathbb{R}^2$ and $z\in E$. Prove that for any $\varepsilon>0$ neigborhood $N_{\varepsilon}(z)$ has point $x\neq z$ such that $x\in E$ 
How to prove this fact strictly?
 A: Orat has given an explicit construction which works in his comment, but it may also be helpful to understand why such a construction is used, along with how you can find similar constructions to solve related problems.
To prove this constructively from first principles, given an arbitrary $x \in E$ and $\varepsilon>0,$ we wish to find a $z$ that satisfies the given properties (and prove that it works). 
Here drawing a picture can be useful to find a general construction - this also applies for more complicated examples. In particular, doing this you will realise that all points on the line segment between $0$ to $x$ is in $E.$ More precisely, for any $\lambda \in [0,1],$
$$ z = \lambda 0 + (1-\lambda)x = (1-\lambda x) \in E $$
Now we want to choose $\lambda$ s.t. $d(x,z)<\varepsilon.$  Here we can expand out the definitions and proceed algebraically (setting $d(x,z)=\varepsilon/2$ helps simplify things), though you can also find a working value through geometric means.
A: If not, $z$ is an interior ponit of $E^{c}$. So it is contradict to $z\in E$.
