A dense set and neighbourhood bases induce a topological basis Looks like I can't get my head around the following proposition.
Let $(T,\mathcal O)$ be a topological space and $S\subseteq T$ a dense set in $T$. If $B(x)$ is a neighbourhood basis of the point $x\in T$, then $\bigcup_{x\in S}B(x)$ is a basis of $\mathcal O$.
In the next part must be a something wrong with my reasoning, maybe you can drop me a bone.
$(\mathbb R,\mathcal O)$ is the topological space induced by the absolute value function ($\mathcal O$ is the family of all open subsets of $\mathbb R$). 
$a\in\mathbb R$
For $x\in\mathbb R$ the set $U_\varepsilon(x)$ is the open ball of $x$ with radius $\varepsilon\in\mathbb R^+$.
The set $M:=\mathbb R\setminus \lbrace a\rbrace$ is dense in $\mathbb R$.
For all $x\in M$: $B(x):=\lbrace U_\varepsilon(x)|\varepsilon\in\mathbb R^+ \wedge \varepsilon\leq|x-a|\rbrace$
$B(a):=\lbrace U_\varepsilon(x)|\varepsilon\in\mathbb R^+\rbrace$
For all $x\in\mathbb R$ the family $B(x)$ is a neighbourhood basis of x.
$C:=\bigcup_{x\in M}B(x)$ must be a basis of $\mathcal O$, but $\mathbb R\in\mathcal O$ cannot be expressed as unions of sets in $C$.
$a\in\mathbb R\wedge a\notin \bigcup_{\gamma\in C}\gamma$ (because $\forall \gamma\in C(a\notin \gamma)$) contradicts that $C$ is a basis of $\mathcal O$.
Thanks for taking the time.
 A: The proposition is false. 
(1).For example let $T=\mathbb R\times [0,\infty),$ let $O$ be the usual topology, and  let $S=\{(x,y)\in T:y>0\}=T$ \ $(\mathbb R\times \{0\}).$ For $p=(x,y)\in S$ let $B(p)=\{D(p,r): 0<r<y\}$ where $D(p,r)=\{(u,v):(u-x)^2+(v-y)^2<r^2\}.$
Then $V=\cup_{p\in S}B(p)$ is not a base for $O.$ We do not even have $\cup V=T,$ but rather $\cup V=S.$
(2).For a 1-dimensional counter-example, let $T=\mathbb R$, let $S=\mathbb Q=\{q_n:n\in \mathbb N\}$ and let $B(q_n)=\{(-2^{-n-m}+q_n,2^{-n-m}+q_n):m\in \mathbb N\}.$ Let $V=\cup_{n\in \mathbb N}B(q_n).$ 
If we let $L(J)$ be the length of a bounded interval $J,$ then $\sum_{J\in V}L(J)=1.$ So any open interval that is a union of members of $V$  cannot have length greater than $1.$ (Also $\cup V\ne \mathbb R.$)
Remark.In a metric space $(X,d),$ if $S$ is dense in $X$ and $B(p)=\{B_d(p,q):q\in \mathbb Q^+\}$ then $\cup_{p\in S}B(p)$ is a base, but we are letting $q$ range over a dense subset of $\mathbb R^+.$
Remark. There are spaces, such as the Sorgenfrey line, that are first-countable and separable but not second-countable. If $T$ is the Sorgenfrey line and $S$ is a countable dense subset of $T,$ and $B(p)$ is a countable local base at $p,$ then $V=\cup_{p\in S}B(p)$ cannot be  a base for $T$ because $V$ is countable and $T$ does not have a countable base.
