Here's a sketch --- see how it goes.
Let $\alpha,\beta$ be in $B$, $\beta\ne0$. First show that ${\alpha\over\beta}=\gamma+\delta,\quad\delta=p+q\sqrt5$ for some $\gamma$ in $B$ and some rationals $p$ and $q$, $0\le p\lt1$, $0\le q\lt1$. Consider $\delta-\epsilon$ for the following five values of $\epsilon$, all of which are in $B$: $0,1,\sqrt5,1+\sqrt5,(1+\sqrt5)/2$. Show that for at least one of these five values of $\epsilon$ the norm of $\delta-\epsilon$ is less than $1$ in absolute value (the norm of $r+s\sqrt5$ is $r^2-5s^2$). Then we have $\alpha=(\gamma+\epsilon)\beta+(\delta-\epsilon)\beta$ and the norm of $(\delta-\epsilon)\beta$ is the norm of $(\delta-\epsilon)$ times the norm of $\beta$, so it's less, in absolute value, than the absolute value of the norm of $\beta$.
EDIT: It's done nicely in Cohn, Advanced Number Theory, pp 108-109 (with some references to earlier pages). I'll summarize.
With $\alpha,\beta$ as above, rationalize the denominator and write ${\alpha\over\beta}={A_1+A_2\omega\over C}$ with $A_1,A_2,C$ integers and $\omega=(1+\sqrt5)/2$. We want to find $\gamma=a+b\omega$ with $a,b$ integers such that $|N((\alpha/\beta)-\gamma)|\lt1$ which is to say we want $|N((A_1/C)-a+((A_2/C)-b)\omega)|\lt1$ Computing this norm, it's $((A_1/C)-a)^2+((A_1/C)-a)((A_2/C)-b)-((A_2/C)-b)^2$ Choose $a,b$, respectively, as the integers closest to $A_1/C,A_2/C$, respectively, and write $P=(A_1/C)-a,\qquad Q=(A_2/C)-b$ Then $-1/2\le P\le1/2,\qquad-1/2\le Q\le1/2$ and we are looking at $f(P,Q)=P^2+PQ-Q^2$ Now you can use calculus to show that $\max|f(P,Q)|=5/16$ given the restriction on $P,Q$, and you're done.