There are no truly incompressible fluids. Even water, which is often treated as incompressible, can undergo compression; it just takes an extraordinarily large pressure to do it and for all practical purposes (other than sound speed) it is incompressible.
That said, the answer to the original question would typically be that the flow is incompressible, which, again, for all practical purposes (other than sound speed) means the fluid is incompressible in that situation as well.
The explanation for why the Mach number actually implies this is fairly simple if a little long-winded. Are you familiar with the Lagrangian and Eulerian frames of reference for fluid flows? The Lagrangian frame of reference is where the frame follows along with a moving fluid element. The Eulerian frame is stationary and sees fluid elements pass by. In the Lagrangian frame, compression can be defined as the change in volume of a fluid element versus a base state, and this quantity can be related back to pressure change using the bulk modulus, E:
\Delta p = -E\dfrac{\Delta \mathcal{V}}{\mathcal{V}_0}.
If this was a solid, E is the equivalent of Young's modulus. In a fluid it is the negative reciprocal of the thermodynamic compressibility.
If you look from the Eulerian frame, the total volume doesn't change and instead you can view compressibility based on the change in density in the region in question and you can relate that to the volume using conservation of mass from some base state (\rho_0,\mathcal{V}_0) as
(\mathcal{V}_0 + \Delta\mathcal{V})(\rho_0 + \Delta \rho) = \mathcal{V}_0\rho_0.
If you expand this expression and take the limit as the \Delta terms go to zero, you get
\dfrac{\Delta\mathcal{V}}{\mathcal{V}_0} = -\dfrac{\Delta\rho}{\rho_0}.
Combining this with our bulk modulus definition results in
\dfrac{\Delta\rho}{\rho_0} = \dfrac{\Delta p}{E}.
Generally speaking, a flow is incompressible when \Delta\rho/\rho_0 \ll 1. From the integrated Euler equation \int dp/\rho + V^2/2 = \text{const.} you can show that the change in pressure is always bounded by the dynamic pressure \rho V^2/2, so we can drop that into our previous equation
\dfrac{\Delta\rho}{\rho_0} \approx \dfrac{\rho V^2}{2E}.
One of the definitions for the speed of sound in a continuum is a^2 = E/\rho, so
\dfrac{\Delta\rho}{\rho_0} \approx \dfrac{V^2}{2a^2} = \dfrac{1}{2}M^2.
In reality, if you are familiar with order, it is probably more accurate to say
\dfrac{\Delta\rho}{\rho_0} = O\left(\dfrac{1}{2}M^2\right).
So what that says is that the relative density change in the flow is directly related to the Mach number of that flow. If M^2 \ll 1, then \Delta\rho/\rho_0 \ll 1. Generally speaking, this usually means that M^2 has to be an order of magnitude less than 1 or smaller, or M^2 < 0.1, or M < 0.316. We usually just simplify that to M<0.3.
In other words, if the Mach number is less than 0.3, the change in density in a given flow is so minute that it can effectively be ignored for most practical purposes. That also implies that for that given situation, the fluid itself is also behaving in an incompressible manner.
