Aside from an algebraic error, you have solved for the critical value of sound speed where the fluid attains Mach 1 -- hence, you have done nothing wrong. The sound speed will assume different values throughout the flow as a function of local temperature. Nevertheless the fluid will attain a condition where the velocity equals a specific sound speed (Mach 1) that depends on the density where the fluid is at rest. There are other constraints to isentropic flow, for example, a maximum attainable velocity which corresponds to the pressure falling to $0$.
At Mach 1 the fluid velocity equals the critical speed of sound $c_*$at local conditions, so $|\nabla\phi|^2 = c_*^2$ and substituting in the Bernoulli equation we have
$$\frac{c^2}{2} + \frac{k\gamma \rho^{\gamma-1}}{\gamma-1} = \frac{c_*^2}{2} + \frac{c_*^2}{\gamma-1} = C \\ \implies c_*^2 \frac{\gamma +1 }{2(\gamma -1)}= C \\ \implies c_*^2 = 2C \frac{\gamma-1}{\gamma+1} \quad (*)$$.
The constant C can be related to the speed of sound $c_0$ when the fluid is at rest. Under such stagnation conditions the fluid velocity is $0$ and, again, using the Bernoulli equation we have
$$ \frac{k\gamma \rho_0^{\gamma-1}}{\gamma-1} = \frac{c_0^2}{\gamma-1} = C $$
Substituting for C in (*) we obtain
$$c_*^2 = 2\frac{c_0^2}{\gamma-1} \frac{\gamma-1}{\gamma+1} = \frac{2c_0^2}{\gamma+1}, $$
which yields
$$\frac{c_*}{c_0} = \sqrt{\frac{2}{\gamma+1}}$$
Typical values for air are $\gamma = 1.4$ and $c_* \approx 0.913 c_0$.
A Streamline (flow line) is a curve with parametric representation $\mathbf{x}_S(s)$ at some fixed time $t$ such that
$$\mathbf{x}_S(s) \times \mathbf{v}(\mathbf{x}_S(s),t) = 0$$
The velocity field is, therefore, tangential to the streamline at any point, and
$$\frac{d \mathbf{x}_S}{ds}(s) = \alpha(s) \mathbf{v}(\mathbf{x}_S(s),t) $$
Defining the quantity $H (\mathbf{x},t) = \frac{\| \mathbf{v}\|^2}{2} + w + \phi_b$, we have
$$\mathbf{v}(\mathbf{x}_S(s),t) \cdot \nabla H(\mathbf{x}_S(s),t) = 0,$$
and, by the chain rule,
$$\frac{\partial}{\partial s} H(\mathbf{x}_S(s),t) = \nabla H(\mathbf{x}_S(s),t)\cdot \frac{d \mathbf{x}_S}{ds}(s) = \alpha(s) \mathbf{v}(\mathbf{x}_S(s),t)\cdot \nabla H(\mathbf{x}_S(s),t) = 0 $$
Therefore, $H$ is constant along a streamline, where value of this constant can be different for each streamline.
What you refer to as the "classical" Bernoulli statement,
$$\tag{1} p + \frac{1}{2} \rho \|\mathbf{v}\|^2 = k,$$
where $k$ is constant at all points in the domain is valid for steady inviscid, incompressible and irrotational flow.
For a proof, begin with the Euler equations governing steady, inviscid flow,
$$\tag{2}\rho (\mathbf{v} \cdot \nabla) \mathbf{v} = - \nabla p,$$
along with the incompressibility condition $\nabla \cdot \mathbf{v} = 0$.
By the general vector identity
$$\nabla (\mathbf{a} \cdot \mathbf{b}) = (\mathbf{a} \cdot \nabla) \mathbf{b} + (\mathbf{b} \cdot \nabla) \mathbf{a} + \mathbf{a} \times (\nabla \times \mathbf{b}) + \mathbf{b} \times (\nabla \times \mathbf{a}),$$
we get with $\mathbf{a} = \mathbf{b} = \mathbf{v}$,
$$\nabla (\|\mathbf{v}\|^2) = \nabla (\mathbf{v} \cdot \mathbf{v}) = 2(\mathbf{v} \cdot \nabla) \mathbf{v} + 2\mathbf{v} \times (\nabla \times \mathbf{v})$$
For irrotational flow, where $\nabla \times \mathbf{v} = 0$, this reduces to
$$\tag{3}(\mathbf{v} \cdot \nabla) \mathbf{v} = \nabla (\frac{1}{2}\|\mathbf{v}\|^2)$$
Since $\rho$ is constant we obtain after substituting (3) into (2),
$$\nabla (\frac{1}{2}\rho\|\mathbf{v}\|^2) = -\nabla p, $$
whence $\nabla (p + \frac{1}{2}\rho\|\mathbf{v}\|^2) = 0$ and (1) must hold where $k$ is a global constant.
Best Answer
By Helmholtz' theorem, any continuously differentiable vector field that vanishes at infinity can be decomposed into irrotational and soleniodal parts of the form
$$\mathbf{u} = \nabla \phi + \nabla \times \mathbf{a}.$$
In potential flow we assume that the velocity field is irrotational, whence
$$\mathbf{u} = \nabla \phi. $$ Applying the continuity condition we have
$$\nabla \cdot \mathbf{u} = \nabla^2 \phi = 0.$$
With suitable boundary conditions there always exists a solution of Laplace's equation for the velocity potential $\phi$ and hence, the velocity field. The Euler equation ensures conservation of momentum and closes the system of equations so we can solve for the pressure field (always).
The Euler equation applies to the general class of inviscid flows (incompressible or compressible) where incompressible potential flow is a special case. With respect to (b), there is no steady, incompressible, inviscid flow (irrotational or rotational) that does not satisfy the Euler equation. Only if the velocity field solves the viscous Navier-Stokes equations will it fail to satisfy the Euler equation.
Also, there is such a thing as unsteady potential flow.