An ellipse is
defined as the set of points, the sum of whose distances from two "focus" points is a constant.
Take the two focus points to be $(c,0)$ and $(-c,0)$, and the sum of the distances to be $2a$. $(x,y)$ is on the ellipse if and only if
$$\sqrt{(x-c)^2+y^2}+\sqrt{(x-(-c))^2+y^2}=2a\tag{1}$$
$$\sqrt{(x-c)^2+y^2}=2a-\sqrt{(x+c)^2+y^2}$$
Square both sides
$$(x-c)^2+y^2=4a^2-4a\sqrt{(x+c)^2+y^2}+(x+c)^2+y^2$$
$$\implies a^2-cx=a\sqrt{(x+c)^2+y^2}$$
Square both sides again and simplify to
$$x^2(c^2-a^2)-a^2y^2=a^2(c^2-a^2)$$
Here we need to assume that $c \neq a$, and we get
$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\tag{2}$$
This is usually simply written
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
where $b=\sqrt{a^2-c^2}$ (since we must clearly choose $a>c$, it follows that $a^2-c^2>0$)
If we define $b$ as the square root as above, then yes, we must clearly have $a>c$, but it is not clear why must we define b this way? Why can't we have $a<c$ and still have an ellipse? The expression $(2)$ followed from the definition of ellipse in $(1)$ and there were no restrictions except possibly that $a>0$.
