This is lemma 5.2.4 from the book introduction to Banach Space theory by Megginson:
Suppose that $X$ is a normed space, that $0<\epsilon<2$, and that $x$ and $y$ are members of $S_X$(the unit sphere) such that $\|x-y\|=\epsilon$. Then there are sequences $(x_n)$ and $(y_n)$ in $S_X$ such that $\|x_n-y_n\|>\epsilon$ for each $n$ and such that $\lim_n x_n=x$ and $\lim_n y_n=y$.
Here is the first part of the proof: It may be assumed that $X$ is a two dimensional real normed space whose underlying vector space is $\mathbb R^2$. Let $x^*$ be a nonzero member of $X^*$ such that $x^*(x-y)=0$. If $x^*x=0$, then it would follow that $x,y\in S_X\cap \ker x^*$ and therefore that $\|x-y\|$ is either $0$ or $2$, a contradiction.
My question is why it can be assumed that $X$ is a two dimensional real normed space? and why $\|x-y\|=0$ or $2$?
The vectors $x$ and $y$ lie on some two-dimensional subspace of $X$, and so we can just work purely in this subspace - i.e. assume $X$ is two dimensional. Similarly, we can just rotate the vectors by some unit modulus complex number to make them real, and work in the real two-dimensional subspace.
Now, in $\mathbb{R}^2$, $S_X$ is a circle, and $\ker x^*$ for nonzero $x^*$ is a line through the origin, so they intersect at exactly two places. $x$ and $y$ lie in the intersection, so their distance is exactly $0$ or $2$.