Reflexive?
$(i_1,j_1)R(i_1,j_1)$. So yes.
Symmetric?
Say $(i_1,j_1)R(i_2,j_2)$ $\epsilon$ $S$ such that $i_1 < i_2$
Then $(i_2,j_2)R(i_1,j_1)$ does not belong to $S$ because $i_2 > i_1$
This can serve as the counter-example.
So, not symmetric.
Anti symmetric?
Say $(i_1,j_1)R(i_2,j_2)$ $\epsilon$ $S$ such that $i_1 < i_2$ As seen above, $(i_2,j_2)R(i_1,j_1)$ won't be true.
Say $(i_1,j_1)R(i_2,j_2)$ $\epsilon$ $S$ such that $((i_1=i_2)∧(j_1<j_2)).$. So, $(i_2,j_2)R(i_1,j_1)$ won't be true because $j_2 > j_1$
Say $(i_1,j_1)R(i_2,j_2)$ $\epsilon$ $S$ such that $((i_1=i_2)∧(j_1=j_2)).$ This is essentially the definition of relexive instance, and hence not a threat to anti-symmetry.
This relation is anti-symmetric.
Since we have $S=ℵ×ℵ$, means depending on the definition and context (0,0) or (1,1) is the least element. This ensures that we can't descend down infinitely. So, well order.
Option C