If $B = B_1 \cup B_2$ where B is open and $B_1$ and $B_2$ are disjoint, then is it true that $B_1$ and $B_2$ are both open? Suppose that $X = \mathbb{R}$ and it equip $X$ with the standard topology on $\mathbb{R}$. If $B = B_1 \cup B_2$ where $B$ is open and $B_1$ and $B_2$ are disjoint, then is it true that $B_1$ and $B_2$ are both open ?
 A: You shouldn't even need a counter-example for this question. For any open set $B$, take ANY subset of $B$, call the subset $B_1$. Then put $B_2 = B \setminus B_1$. From what you are saying, in particular $B_1$ would be open.
So, rephrasing your question: Is it true that ALL subsets of an open set are themselves open? THIS you can answer yourself.
A: Take $B_1=(-1,0)$ and $B_2=[0,1)$. Then $B_1\cup B_2=(-1,1)$ is open, and $B_1$ and $B_2$ are disjoint. But $B_2$ is not open in the standard topology. 
A: the reals are open. the rationals and irrationals are not open. 
A: More general answer (without condition $X=\mathbb R$ equipped with standard topology).
Statement: If $B_1\cup B_2$ is open and $B_1\cap B_2=\varnothing$ then $B_1,B_2$ are both open.
If $X$ is equipped with discrete topology then this is evidently true.
Conversely observe that $X$ (the whole space) is open by definition. If $B\subseteq X$ then $B\cup B^c=X$ and $B\cap B^c=\varnothing$. 
So if it is true then we can conclude that any arbitrary subset of $X$ is open.
Final conclusion: 

The statement is true if and only if $X$ is equipped with the discrete topology (see the comment of copper.hat).

