Show that $(\Bbb R, \tau)$ is not second-countable where $\tau$ defined as: $U \in \tau \iff \{0,1\} \subseteq U \ \text{or} \ U=\emptyset$. Let $\tau$ be a topology on $\Bbb R$ defined as follows:
$$U \in \tau \iff \{0,1\} \subseteq U \ \text{or} \ U=\emptyset.$$
Show that $(\Bbb R, \tau)$ is not second-countable.
Attempt:
Let $\mathcal{B}$ be any basis of $\tau$. We want to show that $\mathcal{B}$ is uncountable.
To this end, notice that since $\mathcal{B}$ is a basis for $\tau$, then by definition, we have:
$$\forall U \in \tau: \forall x \in U. \exists B \in \mathcal{B}.x \in B \subseteq U.\tag1$$
Now, for any $x \in \Bbb R$, define $B_x:=\{0,1,x\}$. Clearly, $B_x \in \tau$. Then, by $(1)$, we have
$$\forall x \in B_x. \exists B_x \in \mathcal{B}. x \in B_x \subseteq B_x.$$
Now, since for any $x \in \Bbb R$, the sets $B_x$ are distinct, we have
$$|\mathcal{B}| \ge |\Bbb R|.$$
Hence, $\mathcal{B}$ is uncountable. Thus, since $\mathcal{B}$ was arbitrary given, we can conclude that $(\Bbb R, \tau)$ is not second-countable. Q.E.D.
Is this correct?
Also, how to show that $(\Bbb R, \tau)$ is separable?
Thanks in advanced.
 A: Yes, your proof for non second-countability works (well, as Jose pointed out there are some issues with the exact way you phrased it, but I think you more or less have the idea of it. You need to pay attention to your quantifiers). Separability follows along the same lines.
Take $A = \{0, 1\}$ which is clearly countable. Now take any $x \in \mathbb{R}$. We wish to show that $x \in \bar A$. So take any open set $U$ such that $x \in U$. Then of course we must have that $\{0, 1\} \subseteq U$, i.e, $ A \cap U \neq \emptyset$. Thus $x \in \bar A$ and $x$ was arbitrary meaning $\bar A = \mathbb{R}$ demonstrating separability.
A: There are several issues with$$(\forall x\in B_x)(\exists B_x\in\mathcal B):x\in B_x\subset B_x.$$In the first place, here $x$ is a fixed real number, and therefore it makes no sense to write $\forall x\in B_x$. In the second place, the conclusion is not that $(\exists B_x\in\mathcal B):x\in B_x\subset B_x$; it should be $(\exists B\in\mathcal B):x\in B\subset B_x$.
You can do it as follows. Let $O_x=\{0,1,x\}$. SInce it is an open set and since $x\in O_x$, there is some $B_x\in\mathcal B$ such that $x\in B_x\subset O_x$. If $x,y\in\Bbb R\setminus\{0,1\}$ and $x\ne y$, then $B_x\ne B_y$, since $x\in B_x$, but $x\notin B_y$. So, $\bigl|\{B_x\mid x\in\Bbb R\setminus\{0,1\}\}\bigr|=|\Bbb R|$, and therefore $|\mathcal B|\geqslant|\Bbb R|$.
Concerning the question about whether or not $(\Bbb R,\tau)$ is separable, that should be posted as another question.
