I have a question about a small part of the proof of the open mapping theorem. I've boxed it in red and posted the proof that I am looking at below. This is from Axler's Measure, Integration & Real Analysis. How is he applying $\frac{r}{\Vert h\Vert} h$ to get 6.82? This is the only part of this proof I'm not understanding, but it looks like it should be obvious. Here $B$ is the unit ball centered at 0, in case you don't want to read the entire proof. I feel like there is some sort of cancellation going on between $\frac{r}{\Vert h\Vert} h$ and $\frac{\Vert h\Vert}{r}B$ but I'm not sure. Any help is appreciated!
Axler shows that $(\star)$ if $h \in W$ and $\lVert h \rVert \le r$ and $\varepsilon >0$ then for some $f \in B$, $\lVert h - Tf\rVert < \varepsilon$.
What you need to show is that if $h \in W$ and $\varepsilon >0$ then for some $f \in \frac{\lVert h \rVert}{r}B$, $\lVert h - Tf\rVert < \varepsilon$. To this end, take $h \in W$ and $\varepsilon >0$. Now note that $\frac{r}{\lVert{h}\rVert} h$ has norm $\le r$. Thus by $(\star)$, we have that there is some $f \in B$ such that $\lVert \frac{r}{\lVert{h}\rVert} h - Tf \rVert < \frac{r\varepsilon}{\lVert h \rVert}$, that is, $\lVert h - T \left( \frac{\lVert h \rVert}{r} f \right) \rVert < \varepsilon$. Notice that $\left( \frac{\lVert h \rVert}{r} f \right) \in \frac{\lVert h \rVert}{r} B$ and you are done.
Recall that if $W$ is any subset of normed vector space $V$ then for any $r>0$, we define $rW$ to the subset of $V$ which contains elements of the form $rw$ for some $w \in W$.