Let $\tau$ be a topology on some set $X$ and $\mathcal B$ a basis of $\tau$. Let $b\in\mathcal B$ be arbitrary. Now, define $R_b$ as the set of all unions of any subset(s) of $\mathcal B\setminus\{b\}$, (it can be thought as some sort of a "spanning set" of the basis not including the basis element we are interested), so
$$R_b=\left\{U\in\tau\mid\exists \mathcal S\subseteq(\mathcal B\setminus\{b\}):U= \bigcup _{v\in \mathcal S} v\right\}$$
I need to prove that there exists some $b'\in\mathcal B$ such that $b'\in R_{b'}$.
I know the definition of a basis but I don't know how to tackle this problem, maybe the fact that for every intersection of two elements of a basis there is another element of the basis contained in it may be of help, but I don't see how. Any help would be appreciated. Thanks.
This is not true in general. Take $\tau$ to be the discrete topology on $X$, with the basis $\mathcal{B} = \{ \{ x \} : x \in X \}$ of all singletons. It is easy to show for each $b = \{ x \} \in \mathcal{B}$ that $R_b = \{ A \subseteq X : x \notin A \}$. So not only is $b$ not an element of $R_b$, we actually have that $b$ is disjoint from every set in $R_b$.
Suppose $\tau$ is a topology on $X$ with a base $\mathcal{B}$ having this property. That is, there is a $b^\prime \in \mathcal{B}$ such that $b^\prime \in R_{b^\prime}$. This means that for each $x \in b^\prime$ there is some other $b_x \in \mathcal{B} \setminus \{ b^\prime \}$ with $x \in b_x \subsetneqq b^\prime$. In particular $b^\prime$ cannot be the smallest open neighbourhood of any of its elements, because $b_x$ is a strictly smaller open neighbourhood of $x \in b$.
And this should work the other way around. If $\mathcal{B}$ is a base for $\tau$ which contains some $b^\prime$ which is not the smallest open neighbourhood of any of its elements, then $b^\prime \in R_{b^\prime}$. (Each $x \in b^\prime$ has an open neighbourhood $V$ with $x \in V \subsetneqq b^\prime$, and so there is a $b_x \in \mathcal{B}$ with $x \in b_x \subseteq V \subsetneqq b^\prime$, meaning $b_x \in \mathcal{B} \setminus \{ b^\prime \}$. Then $\{ b_x : x \in b^\prime \} \subseteq \mathcal{B} \setminus \{ b^\prime \}$ and $\bigcup_{x \in b^\prime} b_x = b^\prime$.)
In particular, if $\tau$ is a topology on $X$ such that no point in $X$ has a smallest open neighbourhood, then every (nonempty) set in any base for $\tau$ will be as desired. (E.g., $\tau$ could be the usual (metric/order) topology on $X = \mathbb{R}$.)