My question reads:
Let $A$ be a set with at least three elements. If $P=\{B_1,B_2\}$ is a partition of $A$, $C_1$ is a partition of $B_1$, $C_2$ is a partition of $B_2$, and $B_1$ does not equal $B_2$, prove that $C_1 \cup C_2$ is a partition of $A$.
I understand how to show $C_1 \cup C_2$ is non-empty, and then either $X=Y$ or they are disjoint, but I do not understand how to prove that $\bigcup_{X \in C_1 \cup C_2} X=A$. As is how to show the union over $X$ is a subset of $A$ and how $A$ is a subset of the union over $X$. I can not seem to figure out how to prove the last portion of the partition definition in this case.
In order to prove that $C_1\cup C_2$ is a partition of $A$ it must be shown that:
Combining $B_1\neq B_2$ with the fact that $P$ is a partition of $A$ we arrive at the conclusion that $B_1,B_2$ are disjoint non-empty subsets of $A$ with $A=B_1\cup B_2$.
For $i=1,2$: if $R\in C_i$ then $R$ is element of a partition of $B_i$ hence $R$ is a non-empty subset of $B_i\subset A$.
This proves the first bulletpoint.
For $i=1,2$: if $x\in B_i$ then $x\in R$ for some $R\in C_i$.
Combined with $A=B_1\cup B_2$ this proves the second bulletpoint.
If $R,S\in C_1\cup C_2$ with $R\neq S$ and $R,S\in C_i$ for some $i\in\{1,2\}$ then $R\cap S=\varnothing$ because we are dealing with distinct elements of a partition. If $R\in C_1\wedge S\in C_2$ then $R\subseteq B_1$ and $S\subseteq B_2$ so that $R\cap S=\varnothing$ is implied by $B_1\cap B_2=\varnothing$. Likewise we find $R\cap S=\varnothing$ if $R\in C_2\wedge S\in C_1$.
This proves the third bulletpoint