For a metric space $(X,d)$ and an infinite cardinal number $\kappa$, the following are equivalent:

- $X$ has a base of cardinality $\le \kappa$.
- X has a network of cardinality $\le \kappa$. (A network is a collection $\mathcal{N}$ of subsets of $X$ such that every open set is a union of elements from $\mathcal{N}$; a base is just a network that consists of open sets.)
- Every open cover of $X$ has a subcover of cardinality $\le \kappa$.
- Every closed discrete subspace $A$ of $X$ has cardinality $\le \kappa$.
- Every discrete subspace $A$ of $X$ has cardinality $\le \kappa$.
- Every pairwise disjoint family of non-empty open sets of $X$ has cardinality $\le \kappa$.
- $X$ has a dense subspace of cardinality $\le \kappa$.

$1)\rightarrow 2)$ is obvious, and true for all topological spaces $X$.

$2)\rightarrow 3)$ is true in general as well:
Let $\mathcal{N}$ be a network with $\left|\mathcal{N}\right| \le \kappa$.
If $\mathcal{U} = \left\{ U_i : i \in I \right\}$ is an open cover of $X$, then for each $x \in X$ we pick $i(x) \in I$ and $N_x \in \mathcal{N}$, such that $x \in N_x \subset U_{i(x)}$.
Then $\left\{N_x : x \in X\right\} = \mathcal{N}'$ has cardinality $\le \kappa$, and for each distinct element $A$ from $\mathcal{N}'$ we pick $U(A)$ from $\mathcal{U}$ with $A \subset U(A)$ ($A = N_x$ for some $x$, and we pick $U(A) = U_{i(x)}$).
Then $\left\{U(A) : A \in \mathcal{N}'\right\}$ is the required subcover.

$3)\rightarrow 4)$ is always true as well: Let $A$ be closed and discrete. Each $x \in A$ has an open neighbourhood $U_x$
that intersects $A$ in $\{x\}$ only. The open cover $\mathcal{U} = \left\{U_x : x \in A\right\} \cup \{X \setminus A\}$
cannot spare any $U_x$ (or $x$ will not be covered), so the cover $\mathcal{U}$ has
cardinality $|A|$ and no subcover of cardinality strictly less than $|A|$.
So $|A| \le \kappa$, or we'd have a contradiction with 3).

$4)\rightarrow 5)$ Here we need only perfect normality of $X$, in the sense only that
each open set is a countable union of closed sets, or equivalently that each closed set is a $G_\delta$. Let $A$ be discrete, then I claim that $A$ is open in $\overline{A}$.

Proof of claim (needs only that singletons are closed): let $x$ be in $A$ and let $U_x$ be an open neighbourhood of $x$ that intersects $A$ only in $\{x\}$. This $U_x$ has the property that $\overline{A} \cap U_x = \{x\}$ as well: $y \neq x$ and $y \in \overline{A} \cap U_x$, then $U_x\setminus\{x\}$ is an open neighbourhood of $y$, $y \in \overline{A}$ so $U_x\setminus\{x\}$ must intersect $A$, but this can only happen in $\{x\}$, contradiction, so that $\{x\}$ is open in $\overline{A}$.

But then, as $A$ is perfectly normal (being metrisable), $A = \cup_{i \in \mathbb{N}} A_i$ where the $A_i$ are closed in $\overline{A}$ (and thus closed in $X)$. So the $A_i$ are closed and discrete, and by 4) we have $|A_i| \le \kappa$. So $|A| \le \aleph_0 \cdot \kappa = \kappa$, as well.

$5)\rightarrow 6)$ is true for all topological spaces: pick $x_i \in U_i$ for any
pairwise disjoint family $\left\{U_i : i \in I\right\}$ of non-empty open sets. By definition we have that $\left\{x_i: i \in I\right\}$ is discrete (as witnessed by the $U_i$), and so $\left|I\right| \le \kappa$, and 6) has been proved.

$6)\rightarrow 7)$ Here we need the metric in a more essential way.
For each $n \in \mathbb{N}$, let $D_n$ be a family of points with the property that
$x,y \in D_n$ with $x \neq y$ implies $d(x,y) \ge \frac{1}{n}$, and $D_n$ is maximal with that property. Here we use Zorn's lemma, or some equivalent principle.
Note that the balls with radius $\frac{1}{2n}$ around the points of $D_n$ are disjoint
so that $|D_n| \le \kappa$ by 6).

Let $D = \cup_n D_n$, we claim that $D$ is dense in $X$.
We already see that $D$ is of the right size, as $|D| \le \aleph_0 \cdot \kappa = \kappa$.
For if $x$ is not in $\overline{D}$, we have that $d(x,\overline{D}) > 0$ and so for some $m \in \mathbb{N}$ we know that $d(x,\overline{D}) > \frac{1}{m}$. But then, for this $m$,
$d(x,\overline{D_m}) \ge d(x,\overline{D}) > \frac{1}{m}$ and in particular: $d(x,y) > \frac{1}{m}$ for all $y \in D_m$. But then we could have added $x$ to $D_m$ and would have obtained a strictly larger $D_m$, and this cannot be. So $D$ is dense.

$7)\rightarrow 1)$ This needs the metric "most". Let $D$ be the dense subset of
cardinality at most $\kappa$. Let $\mathcal{B} = \left\{B(x,r): x \in D; r \in \mathbb{Q}\right\}$, then $\left|\mathcal{B}\right| \le \aleph_0 \cdot \kappa = \kappa$. I claim that $\mathcal{B}$ is a base for $X$: let $U$ be open and $x \in U$. Some $\epsilon>0 $ exists such that $B(x,e) \subset U$, and as $D$ is dense there is some $y \in D$ in
$B(x,\frac{\epsilon}{3})$. Now pick $r \in \mathbb{Q}$ such that $\frac{\epsilon}{3} < r < \frac{\epsilon}{2}$, then $x \in B(y,r)$ (which is from $\mathcal{B}$) and $B(y,r) \subset B(x,\epsilon)$: if for some $z$, $d(z,y) < r$ then $d(z,x) \le d(z,y) + d(y,x) < r + r < \epsilon$, and so there is a $B_x = B(y,r)$ from $\mathcal{B}$ such that $x \in B_x \subset U$, as required for a base.

This concludes the proof of the equivalence, which shows that weight, network weight, Lindelöf number, extent, cellularity and other cardinal invariants are all the same for metrisable spaces.