To prove the inequality $ab \ge 4$ we substitute from the equalities. We so obtain an estimate
$$ab = \left(c + \frac{1}{d}\right)\left(d + \frac{1}{c}\right) = cd + 1 + 1 + \frac{1}{cd} \ge 4,$$
where we use in the last inequality well-known fact that $x + 1/x \ge 2$ holds for all positive reals $x = cd > 0$.

To find the minimum we use similar way. Substitution for $a$ and $b$ yields
$$ab + cd = \left(2 + cd + \frac{1}{cd}\right) + cd = 2 + 2cd + \frac{1}{cd}.$$
Now we use an inequality $x + y \ge 2\sqrt{xy}$ which holds true for any non-negative reals $x, y$. The choice $x = 2cd$, $y = 1/cd$ follows
$$2cd + \frac{1}{cd} \ge 2\sqrt{2}.$$
Now we see that $ab + cd \ge 2(1 + \sqrt{2})$. To prove that it is the desired minimum we find some $a$, $b$, $c$, $d$ such that they makes an equality in the inequality.
The equality comes in the use inequality if and only if $x = y$, it is $2cd = 1/cd$. It is true e.g. for $c = 1$, $d = \sqrt{2}/2$ and for that values we find $a = 1 + \sqrt{2}$, $b = 1 + \sqrt{2}/2$. Such quadruple satisfies the desired equalities and it holds $ab + cd = 2(1 + \sqrt{2})$ too.