Let the elements be $a,b,c,d,e$. Since $ab+ac+bc$, $ab+ad+bd$, $ab+ae+be$ are rational, their differences, $(a+b)(c-d)$ and $(a+b)(d-e)$, are rational. In general, the sum of two elements multiplied by the difference of two others is rational. Also, $\frac{c-d}{d-e}$ is rational. In general, the quotient of any two differences between the numbers is rational. So WLOG set the minimum of the elements to be $e$, and $d=e+r_4m$, $c=e+r_3m$, $b=e+r_2m$, $a=e+r_1m$ with $m$ irrational, $r_1,r_2,r_3,r_4$ rational. We have $(d+e)(b-c)$, or $(e+e+r_4m)(m)(r_2-r_3)$, is rational, so $(2e+r_4m)m$ is rational. Also $(c+e)(a-b)$, or $(e+e+r_3m)(m)(r_1-r_2)$ is rational, so $(2e+r_3m)m$ is rational. Then $(r_3-r_4)m^2$ is rational, so $m^2$ is rational. Then $2em$ is rational, so $em$ is rational. Since $(em)\left(\frac{m}{e}\right)$ is rational, $\frac{e}{m}$ is rational, so we can set $e=r_5m$, $r_5$ rational. Then each of $a,b,c,d,e$ is equal to $m$ multiplied by a rational number, and we're done.