Is discrete topology countable?

Let $$\left[ {X,\tau } \right]$$ be a topological space, then $$X$$ is said to be the first countable space if for every $$x \in X$$ has a countable local base; i.e., if every $$x \in X$$, $${{\rm B}_x}$$ is countable.

In other words, a topological space $$\left[ {X,\tau } \right]$$ is said to be the first countable space if every point $$x$$ of $$X$$ has a countable neighbohood base. A first countable space is also said to be a space satisfying the first axiom of countability.

Example:

If $$X$$ is finite, then $$\left[ {X,\tau } \right]$$ is first countable space. As $$X$$ is finite, all of its subsets are finite. If $${{\rm B}_x}$$ is a local base of $$x \in X$$, then $${{\rm B}_x}$$ is also finite. So, $$\left[ {X,\tau } \right]$$ is the first countable space.

Example:
Let $$X = \left\{ {a,b,c,d} \right\}$$ be a non-empty set and $$\tau = \left\{ {\phi ,X,\left\{ a \right\},\left\{ b \right\},\left\{ {a,b} \right\},\left\{ {c,d} \right\},\left\{ {a,c,d} \right\},\left\{ {b,c,d} \right\}} \right\}$$ be a topology defined on $$X$$.

Base at $$a$$ $$ = {{\rm B}_a} = \left\{ a \right\}$$ Base at $$b$$ $$ = {{\rm B}_b} = \left\{ b \right\}$$ Base at $$c$$ $$ = {{\rm B}_c} = \left\{ {c,d} \right\}$$

Base at $$d$$ $$ = {{\rm B}_d} = \left\{ {c,d} \right\}$$

Here each local base is countable, so $$\left[ {X,\tau } \right]$$ is the first countable space.

Example:

If $$X$$ is either countable or uncountable and $$P\left[ X \right]$$ is a discrete topology on $$X$$, then $$\left[ {X,\tau } \right]$$ is always the first countable space, because for each $$x \in X$$, $${{\rm B}_x}$$ [singleton] is the local base and so the local base is finite. $${{\rm B}_x} = \left\{ {\left\{ x \right\}} \right\}$$, $$x \in {{\rm B}_x} \subseteq U$$ and $${{\rm B}_x}$$ is countable [finite].

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I saw a proof on proofwiki but couldn't understand it. I only know the bare basics of topology [closed, open, limit points, neighbourhoods]. Is there a simple proof for this?

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I know that there are countable topological spaces with the discrete topology, which are not second countable. But I cannot find an argument, why uncountable spaces with the discrete topology should be second countable.