Consider the topological space $(\mathbb{R}, \mathcal{T}:=\{U \subseteq \mathbb{R} \mid \forall x \in U: \exists \epsilon > 0: [x, x + \epsilon[ \subseteq U\})$. Is this space separable? Is it first countable? Second countable?
First of all, it looks a lot like this space is metrisable, but I tried some metrics and couldn't find a metric that induces the topology, so I tried to prove it in the hard way.
Separable:
Let $G$ be a non empty open set. Let $x \in G$ be fixed. Then, there is $\epsilon > 0$ s.t. $[x,x + \epsilon[ \subseteq G$. Invoking the denseness of $\mathbb{Q}$ in $\mathbb{R}$, we can pick $q \in [x, x + \epsilon[ \cap \mathbb{Q}$, and it follows that $\mathbb{Q} \cap G \neq \emptyset$. Hence, $\mathbb{Q}$ is a dense countable set.
Second countable ($\implies$ first countable)
Let $\mathcal{B}:= \{[x,x + \epsilon[ \mid x, \epsilon \in \mathbb{Q}\}$.
First, it seemed reasonable to claim that this is a basis for $\mathcal{T}$.
Indeed, let $x \in B \in \mathcal{T}$. Then, for some $\epsilon > 0$, we have $[x,x+ \epsilon[ \subseteq B$. Choose $y,z \in [x,x + \epsilon[\cap \mathbb{Q}$ with $y < z$. Then $[y,y + (z-y)] \subseteq [x,x + \epsilon[ \subseteq B$. But this doesn't work, as $x \notin [y,y+(z-y)]$ necessarily.
How can I proceed?