Define Z=min(x,y)

Without loss of generality, suppose x,y are supported on the range [0,1].

$\displaystyle P(Z \leq z) = \int_{x=0}^{x=z} \int_{y=x}^{y=1} f_{x,y} ~dx dy ~~ + \int_{y=0}^{y=z} \int_{x=y}^{x=1} f_{x,y} ~dy dx$

interpretation: This is the chance that X is the smallest and less than z, plus the chance that Y is the smallest and less than z
$\displaystyle P(Z \leq z) = \left( \int_{x=0}^{x=z} \int_{y=0}^{y=1} f_{x,y} ~dx dy - \int_{x=0}^{x=z} \int_{y=0}^{y=x} f_{x,y} ~dx dy \right)~~ + \left(\int_{y=0}^{y=z} \int_{x=0}^{x=1} f_{x,y} ~dy dx - \int_{y=0}^{y=z} \int_{x=0}^{x=y} f_{x,y} ~dy dx\right)$

$\displaystyle P(Z \leq z) = \left(F_X(z) - \int_{x=0}^{x=z} \int_{y=0}^{y=x} f_{x,y} ~dx dy \right)~~ + \left(F_Y(z) - \int_{y=0}^{y=z} \int_{x=0}^{x=y} f_{x,y} ~dy dx\right)$

$\displaystyle P(Z \leq z) = \left(F_X(z) + F_Y(z) \right) - \left( \int_{x=0}^{x=z} \int_{y=0}^{y=x} f_{x,y} ~dx dy + \int_{y=0}^{y=z} \int_{x=0}^{x=y} f_{x,y} ~dy dx\right)$

Consider the integrals on the right hand side, and the square with points at (0,0) and (z,z). The square can be divided into two triangles by drawing a line from (0,0) to (z,z). The first integral covers the lower triangle, and the second integral covers the upper triangle. Therefore, together they are integrating over the whole square. In the more general case where the support of each variable was not (0,1), the same argument would apply - except that we would be dividing an arbitrary rectangle instead of a square
$\displaystyle P(Z \leq z) = \left(F_X(z) + F_Y(z) \right) - \left( \int_{x=0}^{x=z} \int_{y=0}^{y=z} f_{x,y} ~dx dy \right) $

$\displaystyle P(Z \leq z) = \left(F_X(z) + F_Y(z) \right) - P(X \leq z,Y \leq z) $