Equality between weight and density in metric spaces I have to prove that in any metric (in generalized version metrizable) space weight of the space is equal to its own density.
My job done so far:
$$(X,\delta)$$
Is topological space with metric given by
$$e(x,y)$$
$$d(x)- \text{density}$$
$$w(x)- \text{weight}$$
Let $$d(x)=\kappa$$
and $$\mathscr D= ( d_{\alpha}: \alpha < \kappa) $$
$D$ is dense subset of X. For every $$d_\alpha$$
we define a family of sets as
$$\mathscr B_\alpha = (B_{1/n}(d_\alpha): n \in N)$$
(I suppose that $N$ means positive natural numbers)
Let
$$\mathcal B = \bigcup_{\alpha < \kappa}\mathscr B_\alpha $$
Than from cardinal numbers algebra we know that
$$|\mathcal B|=|\mathscr D|*|\mathscr B_\alpha|=|\mathscr D|=\kappa$$
From that $$\mathcal B$$ will be countable base of X which we are looking for.
Let $$x \in B_{1/n}(x)$$
(Open ball around $x$ with radius $\frac 1 n$)
I want to show that there exist
$$B_{\epsilon}(d_\alpha) \in \mathcal B$$
such
$$x \in B_{\epsilon}(d_\alpha) \subseteq B_{1/n}(x)$$
We know that there exist such
$$d_\alpha$$
that
$$e(x,d_\alpha)<\epsilon$$
And now I am stuck. I tried a bunch of estimations to pick a right ball around $d_\alpha$ to contain $x$ in it, and not intersect border of $x'$ ball, but somehow I failed. Any help would be appreciated.
Also sorry for my poor English.
 A: Suppose that $x\in\mathscr{D}$ and $y\in B\left(x,\frac1n\right)$. Let $\epsilon=\frac1n-d(x,y)>0$, and let $m\in\Bbb Z^+$ be such that $\frac1m<\epsilon$. $D$ is dense in $X$, so there is a $p\in D$ such that $d(y,p)<\frac1{2m}$. Clearly $y\in B\left(p,\frac1{2m}\right)$, and we’d like to show that $B\left(p,\frac1{2m}\right)\subseteq B\left(x,\frac1n\right)$.
Suppose that $z\in B\left(p,\frac1{2m}\right)$. Then 
$$\begin{align*}
d(z,x)&\le d(z,p)+d(p,x)\\
&\le d(z,p)+d(p,y)+d(y,x)\\
&<\frac1{2m}+\frac1{2m}+d(x,y)\\
&=\frac1m+d(x,y)\\
&<\epsilon+d(x,y)\\
&=\frac1n\;,
\end{align*}$$
so $z\in B\left(x,\frac1n\right)$, as desired. 

There are a couple of relatively minor problems with your proof so far. $\mathscr{D}$ is a set of points, not a sequence of points, so you really should write
$$\mathscr{D}=\{d_\alpha:\alpha<\kappa\}\;.$$
Similarly, you should use curly braces and not parentheses in the definition of $\mathscr{B}_\alpha$:
$$\mathscr{B}_\alpha=\{B_{1/n}(d_\alpha):n\in\Bbb Z^+\}\;.$$
The base $\mathcal{B}$ is is countable only if $\kappa=\omega$; I expect that you meant to say that $\mathcal{B}$ is the base of cardinality $\kappa$ for which we were looking.
The biggest error is when you let $x\in B_{1/n}(x)$: you need to pick an arbitrary point of $B_{1/n}(x)$, not necessarily the centre point $x$. Also, $\mathcal{B}$ contains only $\frac1n$-balls around the poinst of $\mathscr{D}$, not arbitrary $\epsilon$-balls, so you should let $y\in B_{1/n}(x)$ and try to find $d_\alpha\in\mathscr{D}$ and $m\in\Bbb Z^+$ such that 
$$y\in B_{1/m}(d_\alpha)\subseteq B_{1/n}(x)\;.$$
