Do you want a proof of Ampere's Law? Some book really follows the way you said. I think it is just an example rather than a proof.
For the proof of Ampere Law, there is no need to use the delta function, although this method is more simple in my opinion. Some geometry calculation is enough, but it is more tricky to use this method.
L1 is the source current. $P$ is a field point at $\boldsymbol{r}_2$ whose magnetic field we are interested in, then we have, $\boldsymbol{B}(P)$ according to the Biot-Savart law
Then we calculate the line integral along $L_2$ passing through $P$.
$$\boldsymbol{B}(\boldsymbol{r}_2)\cdot\mathrm{d}\boldsymbol{l}_2=\frac{\mu_0I}{4\pi}\oint_\limits{(L_1)} \frac{\mathrm{d}\boldsymbol{l}_2\cdot (\mathrm{d}\boldsymbol{l}_1\times\hat{\boldsymbol{r}}_{12})}{r_{12}^2}=\frac{\mu_0I}{4\pi}\oint_\limits{(L_1)} \frac{(\mathrm{d}\boldsymbol{l}_2\times\mathrm{d}\boldsymbol{l}_1)\cdot\hat{\boldsymbol{r}}_{12}}{r_{12}^2}$$$$=\frac{\mu_0 I}{4\pi}\oint_\limits{(L_1)}\mathrm{d}\omega=\frac{\mu_0 I}{4\pi}\omega$$
Usually it takes at least 20 minutes to make it clear in class. I wish I could tell you the name of the book I used. But unfortunately, it is writen in Chinese.
I present you the main points of the demonstration, and I think it would be clear to you if you are familiar with the integral and vector analysis. Just be clear that the -dl2×dl1 can be treated as the area between the souce L1 and the L1' which is of a small displacement dl2 relative to L1
Ok, now we are calculating $B(\vec{r_2})\cdot{d\vec{l_2}}$, where $d\vec{l_2}$ is a small displacement in the line integral $\oint_{(L_2)}$. Now we have
$$B(\vec{r_2})\cdot{d\vec{l_2}}=\frac{\mu_0}{4\pi}\oint_{(L_1)}\frac{(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)\cdot\hat{\boldsymbol{r}}_{21}}{r_{21}^2}(1)$$
$(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$ is just the area between line segment $-d\boldsymbol{l}_2$ and $d\boldsymbol{l}_1$. So if we consider the line integral in (1),
$$\oint_{(L_1)}(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$$ is the area between two 'circle', $L_1$ and $L_1'$ (see the first figure of my first answer), where $L_1'$ is another circle with a displacement of $-d\boldsymbol{l}_2$ from $L_1$. But don't forget there is also $\hat{r}_{21}\over {r_{21}^2}$ in the line integral which gives the solid angle with respect to point ${\vec{P}}$.
Ok, now we are calculating $B(\vec{r_2})\cdot{d\vec{l_2}}$, where $d\vec{l_2}$ is a small displacement in the line integral $\oint_{(L_2)}$. Now we have
$$B(\vec{r_2})\cdot{d\vec{l_2}}=\frac{\mu_0}{4\pi}\oint_{(L_1)}\frac{(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)\cdot\hat{\boldsymbol{r}}_{21}}{r_{21}^2}(1)$$
$(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$ is just the area between line segment $-d\boldsymbol{l}_2$ and $d\boldsymbol{l}_1$. So if we consider the line integral in (1),
$$\oint_{(L_1)}(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$$ is the area between two 'circle', $L_1$ and $L_1'$ (see the first figure of my first answer), where $L_1'$ is another circle with a displacement of $-d\boldsymbol{l}_2$ from $L_1$. But don't forget there is also $\hat{r}_{21}\over {r_{21}^2}$ in the line integral which gives the solid angle with respect to point ${\vec{P}}$.
Can you understand what I wrote this time? Then there is not much left for us to move on.
Best Answer
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.