Discrete metric, countable basis? 
Give an example of a metric space which does not have a countable basis.

I was thinking of some uncountable set, with a metric which results in an uncountable number of open subsets. Which resulted in this:
$$(\mathbb{R}, d_{\text{discrete}})$$
Where $$d_\text{discrete}(x,y) = 
\begin{cases} 
1 & \text{if } x\not = y\\
0 & \text{if } x=y 
\end{cases}$$
If I understand this metric correctly then the induced topology $\tau = \{ U\subseteq \mathbb{R}: U \text{ is }d_\text{discrete}\text{-open}\}$ would consists of all subsets of $\mathbb{R}$. And then this would include an uncountable number of sets.
Intuition tells me this cannot have a countable basis. But how do I prove this? Would the following be valid? It doesn't feel very formal...
Proof?
Say there is a countable basis $\mathscr{B}$. Then $(\forall U \in \tau)(U = \bigcup_i B_i)$ where $B_i\in \mathscr{B}$.
But since $\mathscr{B}$ is countable, then $U$ must be countable which leads to a contradiction. ($U$ can be uncountable, like $U=[0,1]$)

Your example is fine, but your argument is not: $\Bbb R$ with the usual topology, for instance, is second countable but has uncountably many open sets. Let $\mathscr{B}$ be a base for the discrete topology on $\Bbb R$. For each $x\in\Bbb R$ the set $\{x\}$ is open, so for each $x\in\Bbb R$ there is a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq\{x\}$. Clearly this means that $B_x=\{x\}$, so $\{x\}\in\mathscr{B}$ for each $x\in\Bbb R$.
