I have to prove that If $X$ is a normed space and $y \in B(x,r) $ (where $x \in X$ and $r>0$) then $U(x,r)\cap U(y,\frac{1}{n})\neq \emptyset $.
I have tried to prove playing with the norms but I have not been able to.
P.S: $B(x,r)$ is a close ball and $U(x,r)$ is a open ball.
There is no loss of generality to assume $x=0$. It will simplify the presentation.
We assume also that $n$ is sufficiently large for
$$\tag{1}2nr>1$$ to hold (see remark below).
Let us take $z$ on the line segment $[x,y]$ at distance $\tfrac{1}{2nr}$ of $y$.
More explicitly, let us define
$$\tag{2}z:=(1-\tfrac{1}{2nr})y.$$
(note that the scalar factor in (2) is >0 due to condition (1)).
Let us show that
$$\tag{3}z \in U(0,r)\cap U(y,\frac{1}{n}).$$
As a direct consequence of (2), $\|z\|=(1-\tfrac{1}{2nr})\|y\|\leq (1-\tfrac{1}{2nr})r<r$. Thus $z \in U(0,r)$.
As (1) can be written $y-z=\tfrac{1}{2nr}y$, we deduce that $\|y-z\|=\tfrac{1}{2nr}\|y\|\leq \tfrac{1}{2nr} r=\tfrac{1}{2n}<\tfrac{1}{n}$.
Thus $z \in U(y, \tfrac{1}{n})$, proving (3).
Remark regarding condition (1): it could look restrictive to establish the property for sufficiently large $n$, thus for sufficiently small fractions $1/n$. But property (3) being established for these $n$, it is clearly true as well for all smaller integers $n$ (the associate balls, being largeer will include the smaller ones).
