I read the following and there is something I don't understand about the notation. Why isn't $S_{ij}$ just zero in the below?
Show that if $T_i$ are the components of covariant tensor $T$, then $S_{ij} = T_i T_j - T_j T_i$ are the components of a skew-symmetric covariant tensor $S$.
The skew-symmetry is obvious. From the transformation law for $T$. $$\overline{T}_i \overline{T}_j - \overline{T}_j \overline{T}_i = T_r \frac{\partial x^r}{\partial \overline{x}^i} T_s \frac{\partial x^s}{\partial \overline{x}^j} - T_s \frac{\partial x^s}{\partial \overline{x}^j} T_r \frac{\partial x^r}{\partial \overline{x}^i} = T_r T_s \frac{\partial x^r}{\partial \overline{x}^i} \frac{\partial x^s}{\partial \overline{x}^j} - T_s T_r \frac{\partial x^r}{\partial \overline{x}^i} \frac{\partial x^s}{\partial \overline{x}^j}= (T_r T_s - T_s T_r) \frac{\partial x^r}{\partial \overline{x}^i} \frac{\partial x^s}{\partial \overline{x}^j}$$ or $$\overline{S}_{ij} = S_{rs} \frac{\partial x^r}{\partial \overline{x}^i} \frac{\partial x^s}{\partial \overline{x}^j}$$ which establishes the covariant tensor character of $S$.
Well, it is zero, but the point of your text is to prove that it indeed is a tensor.
A consequence of this is that if it is zero in one coordinate system ($x^{i}$, say) then it will be zero in every other coordinate system (because of the last expression you wrote).