If a topological space $X$ is $E_{2}$ (second-countable), then it's separable. $X$ is separable if there is a dense enumerable subset in $X$.
I need to prove these three facts:
(i) If a topological space $X$ is $E_{2}$ (second-countable), then it's separable.
(ii) Consider in $\mathbb{R}$ the topology $\tau$ of semiopen sets $[a,b)$.Show that $\mathbb{R}$ is separable, with $\mathbb{Q}$ dense, but there isn't enumerable basis.
(iii) Show that if $M$ is separable, then $M$ is $E_{2}$.
Here are my attempts:
(i): Let $x_i \in U_i$ for each $U_i$ element of the basis. Consider the set $A$ of those elements. The closure of $A$ is equal to $X$...?
(ii) If $\mathbb{B}$ is a basis, then $\forall x \in \mathbb{R}, \exists  B_{x} \in \mathbb{B}$ such that $x \in B_{x} \subseteq [x,x+1)$. Then I need to show that $\mathbb{Q}$ is dense, but how?
(iii) Consider the balls with radius $1/n$ with center in a dense enumerable set, which exists by hypothesis. How to prove that these balls will be a basis? If $B(x;\epsilon)$ is an open set then $B(x;\frac{1}{n}) \subset B(x;\epsilon)$, because we can choose $n > \frac{1}{\epsilon}$.
Can someone tell if what I wrote is right and if so show me how to conclude the proofs?
Thanks.
 A: Your attempt at (i) is fine as far as it goes; you just need to finish it. Let $x\in X$ be arbitrary, and let $V$ be any open nbhd of $x$. Then there is some $i\in\Bbb N$ such that $x\in U_i\subseteq V$, and clearly $x_i\in U_i\cap A\subseteq V\cap A$. Thus, every open nbhd of $x$ meets $A$, and therefore $x\in\operatorname{cl}A$. Since $x$ was arbitrary, $\operatorname{cl}A=X$, and $A$ is dense in $X$. (In fact a set $D$ is dense in $X$ if and only if every non-empty open set in $X$ meets $D$; this is a characterization that I usually find more useful than saying that $\operatorname{cl}D=X$, and its proof is just a small generalization of the argument that I just gave.)
In (ii) you should a bit further with the first part: in order to conclude that $\langle\Bbb R,\tau\rangle$ isn’t second countable, you need to show that if $x\ne y$, then $B_x\ne B_y$. This is pretty clear, so it may be sufficient just to point it out explicitly, but that much at least really should be done. Showing that $\Bbb Q$ is dense is easy if you use the characterization that I mentioned above: each non-empty open set contains a set of the form $[a,b)$, which contains an ordinary open interval $(a,b)$, and you already know that every open interval contains a rational.
For (iii) you want to let $D$ be your countable dense set and show that if $U$ is any non-empty open set, and $x\in U$, then there are a $y\in D$ and $n\in\Bbb Z^+$ such that $x\in B\left(y,\frac1n\right)\subseteq U$. You know that $B(x,\epsilon)\subseteq U$ for some $\epsilon>0$, so all will be well if we can guarantee that there are $y\in D$ and $n\in\Bbb Z^+$ such that $x\in B\left(y,\frac1n\right)\subseteq B(x,\epsilon)$. The trick is to choose $n$ large enough so that $\frac2n<\epsilon$. $D$ is dense in $M$, so there is a $y\in D\cap B\left(x,\frac1n\right)$. Use the triangle inequality to show that
$$x\in B\left(y,\frac1n\right)\subseteq B\left(x,\frac2n\right)\subseteq B(x,\epsilon)\subseteq U\;.$$
