Suppose we have a volume $V$ containing a charge distribution defined by $\rho (\textbf{x})$.
The amount of charge $q~(P)$ located at an arbitrary point $P(x_{0},y_{0},z_{0})$ is :
$$q(P)=\int_{x_{0}}^{x_{0}}\int_{y_{0}}^{y_{0}}\int_{z_{0}}^{z_{0}}\rho(\textbf{x}) \,\mathrm dx\,\mathrm dy\mathrm \,dz=0 \,C$$
This means that the charge at each point insde of $\,V$ is equal to zero coulomb, yet $V$ was defined to be a volume containing charges.
Where does this contradiction come from?
The charge density of a point charge $q$ at $\mathbf{r}_0$ is a Dirac delta function:
$$\rho(\mathbf{r})=q\delta^{(3)}(\mathbf{r}-\mathbf{r}_0).$$
Its integral over any volume containing $\mathbf{r}_0$ is $q$, not 0.
However, you didn’t integrate over any volume at all.
The contradiction is that your integral is not covering any volume. The easy answer is that, by definition, the amount of charge $\text dq$ contained in a volume $\text dV$ at location $\mathbf x$ is just $$\text dq=\rho(\mathbf x)\,\text dV$$
However, if you wanted to relate this to the integral you could just look at a cubic element of length $2\delta$ on each side centered on your point, and then look at what happens as $\delta$ approaches $0$. $$q=\int_{x_0-\delta}^{x_0+\delta}\int_{y_0-\delta}^{y_0+\delta}\int_{z_0-\delta}^{z_0+\delta}\rho(\mathbf x)\,\text dz\,\text dy\,\text dx$$ $$\lim_{\delta\to0}\,q=\text dq=\rho(x_0,y_0,z_0)\,\text dV$$
Something to keep in mind that it seems like is tripping you up: $\text dV$ is not a point. It is still a volume. $\text dV$ technically still has an "infinite number of points" contained within it.
