In this online lecture, the professor writes:
$$E_1 \cup E_2 = (\sqcup_{i=1}^n I_j) \cup (\sqcup_{k=1}^n J_k) = \sqcup_{i=1}^n \sqcup_{k=1}^n(I_j \cap J_k)$$
where $I_j, J_k$ are intervals of $\mathcal{R}$, and the $I_j$ are all disjoint from each other and so are the $J_k$. He justifies it by saying "If two intervals are not disjoint, I can write them as a union of disjoint pieces".
I don't follow the second equality. I can't even see that it's true. For example, if we have $$E_1 = [0,1] \cup [2,3]$$ and $$E_2 = [1,2] \cup [3,4]$$ then $$(\sqcup_{i=1}^n I_j) \cup (\sqcup_{k=1}^n J_k) = [0,4]$$ while $$ \sqcup_{i=1}^n \sqcup_{k=1}^n(I_j \cap J_k) = \{1\} \cup \{2\} \cup \{3\}$$
The professor forgot some complements. $E_1^c$ and $E_2^c$ are both finite unions of intervals, not $E_1$ and $E_2$. This means: $$E_1^c = \bigsqcup\limits_{j = 1}^n I_j$$ $$E_2^c = \bigsqcup\limits_{k = 1}^m J_k$$
Now this implies by De Morgan's law:
$$\left(E_1 \cup E_2\right)^c = E_1^c \cap E_2^c = \left(\bigsqcup\limits_{j = 1}^n I_j \right) \cap \left(\bigsqcup\limits_{k = 1}^m J_k\right) =\bigsqcup\limits_{\substack{1 \le j \le n \\ 1 \le k \le m}} I_j \cap J_k$$