Myhill-Nerode Theorem DFA Minimization
It's much better to understand what this marking means. Rather than just follow an algorithm blindly, it's preferable to understand what's going on.

Let us say that two states $q_1,q_2$ are equivalent if $\delta(q_1,w) \in F$ iff  $\delta(q_2,w) \in F$ for any word $w$. The simplest case in which two states are inequivalent is when $q_1 \in F$ but $q_2 \notin F$ (or vice versa). When this happens, we can mark this pair of states as an inequivalent pair of states. This is what happens in Step 2.

In Step 3, we make the following observation. Suppose that $q_1,q_2$ are inequivalent, and that $\delta(r_1,\sigma) = q_1$, $\delta(r_2,\sigma) = q_2$. Then $r_1,r_2$ are also inequivalent (why?). Accordingly, if $r_1,r_2$ is an unmarked pair (not known to be inequivalent) and $\delta(r_1,\sigma),\delta(r_2,\sigma)$ (known to be inequivalent) then we can mark $r_1,r_2$ as inequivalent as well.

You can show inductively that at the end of Step 3, a pair of states is marked if and only if the two states are inequivalent. Therefore we can merge any pair of unmarked states, obtaining in Step 4 a minimal DFA (again, there is something to prove here).