The problem I have with your (phrasing of your) teacher's statement is the concept "discrete current". What does that even mean?
When you look at the usual Biot-Savart law with continuous current, you consider an infinitesimally small line segment. As the size of the line segment becomes smaller, so does the amount of charge that you consider - until in the limit, you consider an amount of charge $dq$ that tends to zero as $dl$ tends to zero.
The only thing different when you have discrete particles is that the charge never tends to zero - it tends to a finite value. But that in no way invalidates the rest of the analysis.
As @leastaction@ said, if you consider the charge "lumpy" (a delta function) rather than continuous, the equations are virtually unchanged.
I really like the proof contained in the paper Derivation of the Biot-Savart Law from Ampere's Law Using the Displacement Current from Robert Buschauer (2013)
It's simple and it fulfills the role of convincing the reader.
Basically the author works with one point charge $q$ situated in origin of Z azis $(0,0,0)$. He supposes a particle moving in Z axis to positive Z values with velocity $v$. He creates a magnetic field line in a arbitrarious circle with $c$ radius, by symmetry, with center in $(0,0,a)$. The angle between any point in the circle and the center of circle starting from origin $(0,0,0)$ is $\alpha$.
Starting point is a part of 4th Maxwell's Equation of electromagnetism, the Ampere-Maxwell Law that consider changing electric flow with time in a area produces magnetic field circulation. This law generates a magnetic force that can be verified using special relativity that in another reference frame it's just a plain electric force.
$$\oint B\, dl = \mu_0\epsilon_0 \; d/dt(\int_A E.dA)$$
In the left side, the solution consists of integrating the $\oint B dl$ in this circle (butterfly net ring). As $B$ is constant by symmetry, we have
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \oint B\, dl = 2\pi c B \qquad\qquad$ (1)
In the right side $\;[\;\mu_0 \epsilon_0 d/dt (\int_A E\; dA \,)\;],\;$ as the surface (butterfly net) we choose a sphere of radius $r$, to ensure that all points have the same value of electric field:
$$ E = q / 4\pi\epsilon_0r^2$$
Let's first calculate the right-hand integral in the right side. We adopted here a slightly different standard in spherical coordinate. Just to remember,the element for integration into spherical coordinates is $\; r^2 \sin \phi \, dr \, d\phi \, dq $
Let $\theta$ (XY axis) vary from $0$ to $2\pi$ and by consider the angle $\phi$ with the vertical (Z axis) from 0 to $\alpha$.
$$\Phi_E = \int_A E\; dA = q/4\pi \epsilon_0 r^2 \int_A dA = q/(4\pi \epsilon_0 r^2) r^2 \int_{0,2\pi} d\theta \int_{0,\alpha} \sin \Phi\; d\Phi = $$
$$q/4\pi \epsilon_0 2\pi ( -\cos \alpha + 1) = q/2\epsilon_0 (1 - cos\alpha)$$
Thus
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Phi_E = \mu_0 q /2 (1 - cos\alpha)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d \Phi_E / dt = - q/2\epsilon_0 d \cos \alpha/dt\qquad$(2)
Putting $\alpha$ as a function of $z$, we have, by the chain rule:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d \cos \alpha/dt = (d \cos\alpha/dz) \; (dz/dt)\qquad$(3)
However as $z$ is decreasing with the motion at velocity $v$, we have
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad dz / dt = -v \qquad $(4)
On the other hand:
$$ \cos \alpha = z / r = z / \sqrt{c^2 + z^2}$$
Using this online tool for derivation:
$d \cos \alpha/dz = c^2/r^3$ where $r = \sqrt{c^2 + z^2}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad 2\pi c B = q \mu_0 /2 v (c^2/r^3)\qquad$ By (1),(2),(3),(4)
$$B = \mu_0 q v c / 4\pi c r^3$$
but $\quad\sin \alpha = c / r\quad$ so we can add $\quad \sin \alpha\; r / c$:
$$B = \mu_0 q v \sin \alpha /4\pi r^2 $$
Vectorizing we have a cross product:
$$B = \mu_0 q \; v\uparrow \times r\uparrow /4\pi r^3$$
In some infinitesimal point we can consider a element of electric current as a point charge, so we can add other charge points by integration (any force is addictive!) for using in real applications. Thus we have in scalar notation:
$$dB = \mu_0 dq \; v \; r \sin \alpha /4\pi r^2$$
Considering $\quad dq = i\;dt\quad$ and $\quad v = ds/dt\quad $, we finally have reached to Biot-Savart law:
$$dB = \mu_0 i \; ds \; r \sin \alpha /4\pi r^2$$
Best Answer
Since there is no magnetic charge term in the Biot-Savart law, it is only correct if Gauss's law for magnetism ($\nabla \cdot \mathbf{B} = 0$) is true and there are no magnetic monopoles. So it makes sense that Gauss's law can be derived from the Biot-Savart law.
However, the Biot-Savart law cannot be derived from the Maxwell-Ampère Law without implicitly assuming Gauss's law. In general, we know this both because of the lack of a magnetic charge term and because as Giorgio pointed out, the curl and divergence of a vector field are independent quantities. The specific problem with the proof you cited is that it assumes that a continuous vector potential $\mathbf{A}$ can be constructed such that $\nabla \times \mathbf{A} = \mathbf{B}$, which is not true if there are magnetic monopoles.