Since $a$, $b$, $c$, $d \ge 0$, we see that
$$(a+b)^2 \ge a^2 + b^2, \quad (b+c)^2 \ge b^2 + c^2, \quad (c+d)^2 \ge c^2 + d^2, \quad (d+a)^2 \ge d^2 + a^2.$$

Therefore it is enough to prove that $(a+b)(b+c)(c+d)(d+a) \le 1$. However, by the Cauchy-Schwartz inequality, we have
$$(a+b)(b+c)(c+d)(d+a) \le \left( \frac{a+b+b+c+c+d+d+a}{4} \right)^4 = 1.$$
Equality holds only when $a+b = b+c = c+d = d+a$ and $ab = bc = cd = da = 0$. Now it is easy to check that equality holds if and only if $a = c = 1$, $b = d = 0$ or $a = c = 0$, $b = d = 1$.