There is no problem with infinite variance regressors. The assumption is made because it results in consistency at the rate of the square root of the sample size. This is standard and common, thus it is presented in this way in most textbooktextbooks.
In the case of an infinite variance regressor, the OLS estimate converges to the truth at a rate faster than $\sqrt{N}$, and is "superconsistent".
Consider the model: $y_i =\beta_0 +\beta_1 x_{1i}+u_i$.
The OLS estimator of $\beta_1$ is $$\hat{\beta_1} =\beta_1 \frac{Var(x)}{Var(x)} + \frac{Cov(x,u)}{Var(x)}$$
Consider the case of a mean-zero regressor for convenience: $$\hat{\beta_1} =\beta_1 +\frac{\frac{1}{N}\sum_{i=1}^Nx_iu_i}{\frac{1}{N}\sum_{i=1}^Nx_i^2} $$
$$\sqrt{N}(\hat{\beta_1} -\beta_1)=\frac{\frac{1}{\sqrt{N}}\sum_{i=1}^Nx_iu_i}{\frac{1}{N}\sum_{i=1}^Nx_i^2} $$
If a CLT applies to the numerator and the LLN applies to the denominator (i.e., the regressor has finite variance), then the OLS estimate converges at the rate of $\sqrt{N}$. If there is an infinite variance regressor, then the denominator in the above expression does not converge. Suppose however, that $\frac{1}{N^2}\sum_{i=1}^Nx_i^2$ converges to a non-zero constant, then,
$$N^{3/2}(\hat{\beta_1} -\beta_1)=\frac{\frac{1}{\sqrt{N}}\sum_{i=1}^Nx_iu_i}{\frac{1}{N^2}\sum_{i=1}^Nx_i^2} $$
Again, a CLT applies to the numerator, and the denominator converges. Thus, the OLS estimate converges at the rate $N^{3/2}$. This is "superconsistency" and a case that most basic textbooks rule out for ease of exposition.