Meaning of Gauss Law
as radius tends to zero, the expression would give infinity

This is not correct. That a limit is defined does not mean that an expression is defined at the limit. In fact, when the limit is a real number (like 0), why would we even need limits, were this not so?

The ordinary limit of this expression is not defined:

$$\lim_{r \to 0} \frac{Q}{4\pi r^2} = \text{undefined}$$

See division by zero.

However, the one sided limit is defined:

$$ \lim_{r \searrow 0} \frac{Q}{4\pi r^2} = \infty $$

but this still does not mean you can do this:

$$\require{cancel} \cancel{\frac{Q}{4\pi0^2} = \infty} $$

Why do we get infinite field at center?

You don't. The one-sided limit means \$D\$ can be made arbitrarily large by making \$r\$ arbitrarily small but not zero. When \$r=0\$, \$D\$ is undefined, any way you slice it. Using a limit doesn't make dividing by zero defined.

There is a reason \$D\$ is undefined when \$r=0\$: two things can not occupy the same space at the same time. As some other like charge approaches the charge at the center of your sphere, it experiences an increasing repulsive force. There is no limit to how large this force can become, so you can never get the two charges in the same space.

Two things can happen, if you try. The common case is that the charges approach each other until the repulsive force grows to be sufficient to balance whatever other force is pushing the charges closer. This is what prevents you from pushing your finger through the desk, for example.

The less common case is that you get close enough that other forces become significant, and your charge at the center can no longer be considered a point, but must be considered as separate, more fundamental particles. This usually doesn't happen because it requires so much energy. At this point, probably you are playing with a particle accelerator, and you have exited the field of electrical engineering and entered the field of particle physics. I'm not a physicist so I can't tell you exactly what will happen, other than it will look something like this:



I really have no idea what this picture represents, other than whatever was in the middle blew up. I've yet to observe such effects in my circuits, so I haven't had reason to care :)