Trouble comparing the K-topology and Upper Limit topology on $\mathbb{R}$ I don't quite understand why the Upper Limit Topology (ULT) on $\mathbb{R}$ is strictly finer than the K-topology. This solution states that the problem should be solved by using lemma 13.3. Could anyone care to give a hint/nudge towards solving this? I'm an undergrad that is completely new to topology and I'm currently self studying Munkres, so forgive me if the question is trivial.
Lemma 13.3:
Let $\mathcal{B}$ and $\mathcal{B}^{'}$ be bases for topologies $T$ and $T^{'}$ respectively, on $X$. Now $T \subset T^{'}$ iff for each $x \in X$ and each basis element $B \in \mathcal{B}$ containing $x$, there is a basis element $B^{'} \in \mathcal{B}^{'}$ s.t $x \in B^{'} \subset B$
The ULT has the basis of all sets $(a,b]$ and the K-topology has the basis of all sets $(a,b)$ and $(a,b) - K$ where $K$ is the set of all numbers of the form $1/n$, for $n \in \mathbb{Z}_+$
 A: (To make it answered)
@logarithm comment:

$K$-neighborhoods of $x\neq 0$ contain some $(a,b)\ni x$ because $x$ is at positive distance from $K$. Therefore, they contain the ULT-open $(a,x]$. Neighborhooods of $0$ contain some $(a,0]$ since $K$ doesn't contain any negative numbers or zero. The inclusion is strict because $(-2,-1]$ is open in ULT but not in $K$-topology

A: $(\bullet ).$ Let $P,Q$ be  topologies on a set $S.$ If there is a base (basis) $B$ for $P$ such that $B\subset Q$ then $P\subset Q.$
On $\Bbb R$ let $T(0)$ be the usual topology and let $B(0)$ be the base for $T(0)$ consisting of all $(a,b).$  Let $T(1)$ be the $K$-topology and let $B(1)$ be the base for $T(1)$ as in your Q. Let $T(2)$ be the upper-limit topology (a.k.a. the Sorgenfrey topology, or the Sorgenfrey line).
(1).We have $(a,b)=\cup_{c\in (a,b)}(a,c] \in T(2).$ So  $B(0)\subset T(2)$ so  by $(\bullet)$ we have $$T(0)\subset T(2).$$
(2).We have $(a,b)-K=C\cup D\cup E 
\;$ where $$C=\cup_{n\in \Bbb N}\;(a,b)\cap (\frac {1}{n+1},\frac {1}{n}),$$ $$D=(a,b)\cap (1,\infty),$$ $$  E=(a,b)\cap (-\infty,0].$$
(2-i).  $C$ and $D$ belong to $T(0)$ so $C$ and $D $ belong to $T(2)$ by (1).
(2-ii).If  $0\not \in (a,b)$ then $E=(a,b)\cap (\infty,0)\in T(0)\subset T(2)\;$  by (1) so  $E\in T(2).$
(2-ii'). If $0\in (a,b)$ then $a<0<b$ so $E=(a,0]\in T(2). $
So  we have $C,D,E\in T(2).$ So $(a,b)-K=C\cup D\cup E\in T(2).$
So $B(1)\subset T(2).$ So by $(\bullet) $ we have  $$T(1)\subset T(2).$$ 
(3-i).  $0$ is not in the $T(2)$-closure of $\Bbb R^+$ because $0\in (-1,0]\in T(2)$ and $(-1,0]$ is disjoint from $\Bbb R^+$.
(3-ii). If $0\in U\in T(1)$ then $U\supset (a,b)-K$ for some $a,b$ with  $a<0<b,$ so $U\cap \Bbb R^+\supset (0,b)-K\ne \emptyset.\;$ So $0$ is in the $T(1)$-closure of $\Bbb R^+.$
So $T(1)\ne T(2).$ 
Therefore $T(1)\subsetneqq T(2).$ 
Remark. $(\bullet)$ can be re-phrased as "iff", and then it is a special case of Lemma 13.3 when the base for the larger topology is the whole of the larger topology. Sometimes it is easier to work with an entire toplogy rather than a base for it, as you are not bothered by whether or not a union of some members of the base is a member of the base.
