Why is the Laplacian of $1/r$ a Dirac delta? How does one show that $\nabla^2 1/r$ (in spherical coords) is the Dirac delta function ? Intuitively, it would seem that the function undefined at the origin and I'm not able to construct a limiting argument that avoids this problem.

Delta function is zero everywhere except at origin, and integration over space is zero. The first property is easy to prove with vector identities. For the second property:
$$
I = \iiint_V \nabla^2 \frac{1}{r} d V = \iiint_V \nabla \cdot \nabla \frac{1}{r} dV
$$
With divergence theorem:
$$
I = \oint_S \nabla \frac{1}{r} \vec{dS} = \oint_S - \frac{1}{r^2} \vec{n} \vec{dS} = \int_{\Omega} - \frac{1}{r^2} r^2 {d \Omega} = -4 \pi = -4 \pi \iiint_V \delta(r) dV
$$
That's all, your question missed the factor $-4 \pi$.
As to the test function (as response to comment below), the spherical surface in the second integral has to be infinitely small around the origin. In this way $f(r)$ in an integration like (The volume $V$ can be any size but only the infinitesimal region around the origin matters, as at outside laplacian of 1/r is zero):
$$
\iiint_V f \nabla^2 \frac{1}{r} dV
$$
can be replaced by the constant $f(0)$. Of course $f$ must be continuous and finite around the origin.
