question on Banach Limits from Conway's book

Show that if $x\in l^{\infty}$, $||x||_{\infty}\leq1$, then there is a sequence ${x_{n}}$, $x_{n}\in l^{\infty}$ such that $||x_{n}||_{\infty}\leq1$,$||x_{n}-x||_{\infty}\rightarrow 0$ and each $x_{n}$ takes on only a finite number of values.

This is one of the homework problems for my functional analysis course. It's actually in Conway's functional analysis book under the chapter Banach limit page 83. I don't know how to approach using Banach limit Dfinition.

This has nothing to do with Banach limits.

First it is easy to reduce the problem to the case where $x\geq0$ (you first decompose $x$ in its real and imaginary parts, then you write these as the difference of their positive and negative parts).

Now, for each $n$, define $$ x_n(k)=\frac{\lfloor n x_k\rfloor}{n}. $$ What this achieves is to divide $[0,1]$ in $n$ slots, and for each entry one chooses the "closest slot".

By construction, $\|x-x_n\|_\infty\leq\frac1n.$

And $x_n$ only takes the values $0,\frac1n,\frac2n,\ldots,1$.