Let $$ K \colon= \left\{\ \frac{1}{n} \ \colon \ n \in \mathbb{N} \ \right\},$$ and let the $K$-topology on $\mathbb{R}$ be the one having as basis all open intervals $(a,b)$ and all sets of the form $(a,b)-K$, where $a, b \in \mathbb{R}$ such that $a< b$.
Then this topology is finer than the usual topology on $\mathbb{R}$.
And, these two topologies are different.
But what sets are open in the $K$-topology but not in the usual topology on $\mathbb{R}$?
And, how are the subspace topologies that $(0, +\infty)$ inherits as a subspace of $\mathbb{R}$ under these two topologies, respectively, the same, as Munkres claims?
As per my understanding, any set of the form $(a,b)-K$ is again a union of one or more open intervals and is therefore in the usual topology.
Sets of the form $(a,b)-K$ where $a<0< b$ are not open in the usual topology. This is because there is no open interval containing $0$ that does not contain an element of $K$, and given any open set $U$ in the usual topology and any point $x$ in the set there must be an open interval $I$ such that $x\in I\subset U$.
To answer the second question, the reason the subspace topology on $(0,\infty)$ is the same as the usual topology is that the intersection of $K$ with this subset is closed. All of the new open sets in the $K$ topology contain 0.