Note that $T$ is a linear map that is bounded. The only part I am a little confused about is the construction of $\{y_n\}$. I understand everything after that. Note, $0 < \epsilon < 1$. I know that from assumption, $B(0, r) \subset \overline{T(B(0,1))}$. If $y \in B(0, r)$, then $y \in \overline{T(B(0,1))}$. Hence, $$T(B(0,1)) \cap B(y, s) \neq \emptyset$$for all $s > 0$. Namely setting $s := \epsilon \cdot r$. Then, There is $y_1 \in T(B(0,1))$ such that $||y - y_1|| < \epsilon \cdot r$. When he is saying that "By homegeneity" is he using that to show that $\epsilon T(B(0, 1))$ is dense in $B(0, \epsilon r)$? If so, what would be the proof of that?
Put $\tilde{y}=\varepsilon^{-1}(y-y_1)$, then $\tilde{y}\in B(0,r)$ and we find $\tilde{y_2}\in \overline{T(B(0,1))}$ such that $$\Vert \tilde{y}-\tilde{y_2}\Vert <\varepsilon r.$$ Thus, after multiplying the whole thing with $\varepsilon$ we get $$\Vert y-y_1-\varepsilon \tilde{y_2}\Vert =\Vert \varepsilon \tilde{y}- \varepsilon \tilde{y_2}\Vert <\varepsilon^2 r.$$ Now set $y_2=\varepsilon \tilde{y_2}$ and note that $y_2\in \varepsilon \overline{T(B(0,1))}$, which is exactly what is claimed.