The Sorgenfrey line: $\mathbb{R}$ with the half-open intervals $[a,b)$ as a basis is strictly finer than the usual topology, is first countable, has a countable dense subset, and its sequences converge only from the right

$\mathcal{B}$ covers $\mathbb{R}$: for $x\in \mathbb{R}$ one has $x<x+1$ by [L4], so $[x,x+1)\in \mathcal{B}$ and $x\in [x,x+1)$.

Let $x\in [a,b)\cap [c,d)$ and put $a^{\mathrm{\prime }}:=\max \{a,c\}$ and $b^{\mathrm{\prime }}:=\min \{b,d\}$, which exist by [L4]. Then $[a,b)\cap [c,d)=[a^{\mathrm{\prime }},b^{\mathrm{\prime }})$, since $a\le t$ and $c\le t$ together say $a^{\mathrm{\prime }}\le t$ and $t<b$ with $t<d$ says $t<b^{\mathrm{\prime }}$; and $a^{\mathrm{\prime }}\le x<b^{\mathrm{\prime }}$ gives $a^{\mathrm{\prime }}<b^{\mathrm{\prime }}$, so $[a^{\mathrm{\prime }},b^{\mathrm{\prime }})\in \mathcal{B}$ and $x\in [a^{\mathrm{\prime }},b^{\mathrm{\prime }})\subseteq [a,b)\cap [c,d)$.

For $x\in \mathbb{R}$ and real $r>0$: $x<x+r$ by [L4], and $[x,x+r)\subseteq (x-r,x+r)$, since $x-r<x\le t<x+r$.

For every $k\in \mathbb{N}$ the real $1\mathrm{/}(k+1)$ is positive by [L3], so $[x,x+1\mathrm{/}(k+1))\in \mathcal{B}$ by [L4] and contains $x$.

$[0,1)\in \mathcal{B}$, since $0<1$ by [L4].

Every nonempty member $[a,b)$ of $\mathcal{B}$ meets $\mathbb{Q}$: by [L5] there is a rational $q$ with $a<q<b$, and then $q\in [a,b)$.

The same sequence converges to $x$ in the usual topology: given $r>0$, [L3] gives $n\ge 1$ with $1\mathrm{/}n<r$ and $n=m+1$; for $k\ge m$ the canonical naturals satisfy $0<(m+1)\cdot 1_{\mathbb{R}}\le (k+1)\cdot 1_{\mathbb{R}}$, so $1\mathrm{/}(k+1)\le 1\mathrm{/}(m+1)<r$ by [L3], and $\mathrm{\mid }{y}_k-x\mathrm{\mid }=1\mathrm{/}(k+1)<r$, that is $y_k\in B(x,r)$.

By steps 1.1 and 1.2 the family $\mathcal{B}$ satisfies the two basis conditions of [L1], so it is a basis for the topology ${\mathcal{T}}_{\mathrm{S}}$ described there; this is claim 1.

$[0,1)$ is not open in the usual topology: for any $r>0$, [L3] gives a natural $n\ge 1$ with $1\mathrm{/}n<r$, and $-1\mathrm{/}n$ satisfies $-r<-1\mathrm{/}n<0$, so $-1\mathrm{/}n\in (-r,r)$ while $-1\mathrm{/}n\mathrm{\notin }[0,1)$; hence no ball around $0$ lies inside $[0,1)$.

The family $\{[x,x+1\mathrm{/}(k+1)):k\in \mathbb{N}\}$ is nonempty and is the image of the surjection $k\mapsto [x,x+1\mathrm{/}(k+1))$ from $\mathbb{N}$, hence at most countable.

By step 1.6 the set $\mathbb{Q}$ meets every nonempty basic open set, so it is dense by [L6]; with [L5] it is at most countable, which is claim 4.

Every set $U$ open in the usual topology lies in ${\mathcal{T}}_{\mathrm{S}}$: for $x\in U$ take $r>0$ with $(x-r,x+r)\subseteq U$, and then $x\in [x,x+r)\subseteq U$ by step 1.3, with $[x,x+r)\in \mathcal{B}$.

Let $N$ be a neighbourhood of $x$ in ${\mathcal{T}}_{\mathrm{S}}$ and take $[a,b)\in \mathcal{B}$ with $x\in [a,b)\subseteq N$, so $a\le x<b$ and $b-x>0$; by [L3] fix a natural $n\ge 1$ with $1\mathrm{/}n<b-x$ and write $n=m+1$ with $m\in \mathbb{N}$. Then $x+1\mathrm{/}(m+1)<b$, so $[x,x+1\mathrm{/}(m+1))\subseteq [x,b)\subseteq [a,b)\subseteq N$.

For every real $\epsilon >0$ the set $[x,x+\epsilon )$ is a member of $\mathcal{B}$ containing $x$, hence a neighbourhood of $x$ in ${\mathcal{T}}_{\mathrm{S}}$.

By steps 3.1 and 2.2 the topology ${\mathcal{T}}_{\mathrm{S}}$ contains the usual topology and contains $[0,1)$, which the usual topology does not; so ${\mathcal{T}}_{\mathrm{S}}$ is strictly finer, which is claim 2.

By steps 1.4, 3.2 and 2.3 the family of claim 3 consists of neighbourhoods of $x$, is at most countable, and has a member inside every neighbourhood of $x$; so it is an at most countable neighbourhood base at $x$, and $x$ was arbitrary. This is claim 3.

If $x_k\to x$ in ${\mathcal{T}}_{\mathrm{S}}$ and $\epsilon >0$, then by step 3.3 the set $[x,x+\epsilon )$ is a neighbourhood of $x$, so there is $K$ with $x_k\in [x,x+\epsilon )$, that is $x\le x_k<x+\epsilon$, for all $k\ge K$.

Conversely, assume the $\epsilon$ condition and let $N$ be a neighbourhood of $x$; take $[a,b)\in \mathcal{B}$ with $x\in [a,b)\subseteq N$ and apply the condition with $\epsilon :=b-x>0$, obtaining $K$ with $x\le x_k<b$ for all $k\ge K$; since $a\le x\le x_k$, this gives $x_k\in [a,b)\subseteq N$ for all $k\ge K$. So $x_k\to x$.

The sequence $y_k=x-1\mathrm{/}(k+1)$ satisfies $y_k<x$ for every $k$, since $1\mathrm{/}(k+1)>0$; so no term lies in $[x,x+1)$, and by step 4.3 with $\epsilon =1$ the sequence does not converge to $x$ in ${\mathcal{T}}_{\mathrm{S}}$.

Steps 4.3 and 4.4 give the equivalence of claim 5, and steps 5.1 and 1.7 give the sequence it names; with steps 4.1, 4.2, 2.4 and 2.1 all five claims are proved.