Question about basis of a topology The Royden's definition of basis of a topology is exactly as the following.

Definition For a topological space $(X, \tau)$ and a point $x$ in $X$, a collection of neighborhoods of $x$, $B_x$, is called a base for the topology at $x$ provided for any neighborhood $U$ of $x$, there is a set $B$ in the collection $B_x$, for which $B \subseteq U$. A collection of open sets $B$ is called a base for the topology $\tau$ provided it contains a base for the topology at each point.

Let us say that we have an open set $A\in \tau\setminus \{\emptyset\}$, i.e., $A$ is a nonempty open subset of $X$, and consider $\{B_n\}$ a countable basis for the topolgy. 
I want to understand why it is true that $A$ contains some $B_n$ of $\{B_n\}$.
My attempt is: $\exists x \in A$ (by axiom of choice, since $A$ is a nonempty subset of $X$), and by the definition above $\exists \beta \subseteq \{B_n\}$ which is a base for the topology at $x$. Since $A$ is a neighboorhood of $x$, then $\exists H \in \beta: H\subseteq A$. I cannot go further than this. Maybe because I didn't understand precisely the definition above, and (as a consequence) the definitions of first countability and second countability.
Can you help me to understand what am I missing in these concepts?
 A: Part of the problem is that there are too many things called $B$, and the informal notation $\{B_n\}$ also seems confusing.  I'm going to use capital letters for subsets of $X$, and calligraphic letters for sets ("collections") of subsets of $X$.
Suppose then that $\mathcal{B}$ is a base for $(X, \tau)$ (you can think of $\mathcal{B}$ as being countable if you like, but it's irrelevant to this proof).  Since $A$ is not empty, it contains at least one point; let $x$ be such a point*.  Then since $\mathcal{B}$ is a base for $(X, \tau)$, by definition it has a subset $\mathcal{B}_x \subseteq \mathcal{B}$ which is a base at $x$.  By definition of a base at $x$, since $A$ is an open set containing $x$, there exists $U \in \mathcal{B}_x$ with $x \in U$ and $U \subseteq A$.  In particular, $U \in \mathcal{B}$ so we are done.
* This step does not require the axiom of choice.  It is an example of existential instantiation which is valid in first-order logic without any axioms whatever.  You can use it multiple times to get any finite number of elements, if you like.  The axiom of choice is only needed when you need to choose an infinite number of elements from an infinite number of sets; you can't just use existential instantiation an infinite number of times, because a proof is only allowed to contain a finite number of steps.
A: Rather more like this:   $\forall x\in A,\,\exists B_x$ such that $(~\!x\in B_x~\!)\land (~\!B_x\subset A~\!)$.
Then, further, $A=\bigcup_{x\in A}B_x$.  Thus $A$ is a union of basis elements.   This, as far as I know, is the idea of a basis (this author is calling it a "base";  either way).
Then as for first countable, it means each point has a countable (neighborhood) base.
And a topology is second countable if there is a countable basis (base) for it.
A: Since $A$ is a nonempty subset of $X$, $\exists x \in A$ (by axiom of choice, ). The definition above of basis of topology can be rephrased as  $\forall A\in \tau:\forall x\in A: \exists B_x  \in \{B_n\}: x\in B_x \subseteq A$. So $A\supseteq B_x:B_x \in \{B_n\}$. Indeed, $A$ must contain some $B_x$.
Note that a base $C$ for the topology has an element cointained not only in the neighborhoods of a single $x$, but in the neighborhoods of all $x\in X$. I think this was the key element to understand the relation between the definitions of basis for topology at a point, and basis for topology. To be honest, I didn't like the way the author presented both concepts.
