As far as I can tell, this "argument" is not a meaningful derivation of $\mathbf E = -\nabla \Phi$. In step 4, one seems to assume that $Edx = -d\phi$ which is essentially what one is trying to prove in the first place, so the argument seems circular.
The precise derivation of this fact relies on the crucial fact that the electric field is a conservative vector field which then immediately guarantees the existence of a scalar function $\Phi$ for which $\mathbf E = -\nabla\Phi$.
As for your assertion
if electric field is a constant, electric potential must be a constant too.
Notice that if the electric field is constant in some region then the gradient of $\Phi$ is a constant there. In particular its partial derivatives satisfy
\partial_x\Phi = -E_x, \qquad \partial_y\Phi = -E_y, \qquad \partial_z\Phi = -E_z
The $x$ equation gives
\Phi(x,y,z) = -E_x x + f(y,z)
Then the $y$ equation gives $\partial_yf = -E_y$ which implies $f(y,z) = -E_y y + g(z)$ and finally the $z$ equation gives $\partial_z g = -E_z$ so that $g(z) = -E_z z + c$. Putting this all together gives
\Phi(x,y,z) = -\mathbf E \cdot \mathbf x + c
In particular, the electric field being constant does not imply that the electric potential is constant, it means that it has the general form written in this last equation.