Proving that a set is a basis for the neighborhoods 
Take $\mathbb{R}$ with the usual topology $\mathcal{T}$. 
Let $A = \{n^{-1}\mid n \in \mathbb{N}\}$.
Define the topology $\mathcal{S}$ as the topology generated by
  $\mathcal{T} \cup \{\mathbb{R}\setminus A\}$.
Define $\mathcal{B}(x)$ as the set $\{]x - \epsilon,x + \epsilon[:
 \epsilon >0\}$ if $x\neq 0$ and as the set
   $\{]-\epsilon,\epsilon[\setminus A : \epsilon > 0\}$
Prove that this is a basis for the neighborhoods on the topological
  space $(\mathbb{R}, \mathcal{S})$.

My attempt:
I was able to prove that $\mathcal{B}(x)$ is contained in the neighborhoods of the point $x$, but I was not able to formally show that we can write $B \subset V$ for all neighborhoods $V$ of $x$ and for some $B \in \mathcal{B}(x)$. This seems tedious, and I wonder how I can write it neatly. 
 A: If we have $\mathcal{S}' = \mathcal{T} \cup \{(X\setminus A)\}$ as a subbase (a generating set), then we have a base $\mathcal{B}$ consisting of all finite intersections from $\mathcal{S}'$. Such a finite intersection is always of the form $O$ where $O \in \mathcal{T}$ (when the set $X\setminus A$ is not part of the intersection), as then all sets from the intersection come from $\mathcal{T}$, which is closed under finite intersections, or of the form $O \setminus A$, with $O \in \mathcal{T}$ (which includes $X\setminus O$ itself).
So $\mathcal{B} = \mathcal{T} \cup \{O \setminus A: O \in \mathcal{T}\}$ forms a base for the topology $\mathcal{S}$.

(1) For every set $B \in \mathcal{B}$ and every $x \in B$ we can find a $B' \in \mathcal{B}(x)$ such that $x \in B' \subseteq B$, 

and as also $\cup_{x \in X} \mathcal{B}(x) \subseteq \mathcal{S}$ we have that the collection you gave is a family of local bases for $\mathcal{S}$.  
Statement (1) can be shown as follows: suppose $x \in B$ and $B$ is of the first form, i.e. a member of $\mathcal{T}$. Then by properties of the usual topology we have some $\varepsilon >0$ such that $(x-\varepsilon, x+\varepsilon) \subseteq B$, and so we have our member $B' \in \mathcal{B}(x)$ for $x \neq 0$ and for $x \neq 0$ we can take the even smaller set $(-\varepsilon,\varepsilon)\setminus A \in \mathcal{B}(0)$. If $B$ is of the second form, so $B=O \setminus A$ where $O$ open in $\mathcal{T}$ then if $x=0$ is in $B=O\setminus A$ we can find $\varepsilon>0$ such that $(-\varepsilon,\varepsilon) \subseteq O$ and then $\mathcal{B}(0) \ni (-\varepsilon,\varepsilon)\setminus A  \subseteq B$, as required. And if $x \neq 0$ lies in $B$, then $C=A \cup\{0\}$ is closed in $\mathcal{T}$ and $x \in O \setminus C  = O \cap (X\setminus C)$, which is an open set of $\mathcal{T}$, so we have $\varepsilon >0$ again with $\mathcal{B}(x) \ni (x-\varepsilon, x+\varepsilon) \subseteq O \setminus C \subset B$, as required.
