Remember that the open sets are arbitrary unions of elements of $B$, not just elements of $B$. So while it's pretty easy to see that in all of your examples there do not exist $U,V\in B$ such that $U\cup V=\mathbb{R}$ and $U\cap V=\emptyset$, that doesn't mean there does not exist a separation. Arbitrary unions of elements of $B$ can be pretty complicated!
In fact, your first example is not connected. Every subset of $\mathbb{R}$ is open in this topology, since every set is a union of singleton sets (why?). Can you use this to find a separation for this topology?
Answer:
Take $U=\{0\}$ and $V=\mathbb{R}\setminus\{0\}$. Then $U\cup V=\mathbb{R}$, $U\cap V=\emptyset$, $U$ and $V$ are nonempty, and $U$ and $V$ are open, so they are a separation. More generally, let $U\subset\mathbb{R}$ be any nonempty proper subset of $\mathbb{R}$ and let $V=\mathbb{R}\setminus U$.
Your second example happens to give a connected topology, but this is very nonobvious and requires a lot of cleverness to prove (the proof uses the completeness of $\mathbb{R}$). So I wouldn't worry about this example for now.
Your third example also gives a connected topology, and this time it is actually not too hard to prove it. See if you can prove it! As a first step, you need to understand what open sets look like, remembering that an open set is a union of elements of $B$, not just a single element of $B$. What can you say about sets that are unions of elements of $B$? (Hint: You might try proving that if $U$ and $V$ are nonempty and both unions of elements of $B$, then they cannot be disjoint.)
Answer:
Every element of $B$ contains $0$ as an element, and every nonempty open set contains an element of $B$ as a subset. So if $U$ and $V$ are nonempty open sets, $0\in U$ and $0\in V$ so $U\cap V\neq\emptyset$. So there cannot exist any separation. (In fact, every open set is actually equal to a single element of $B$ in this case, but this is harder to prove.)
