The proof I am reading contains the following statement:
... we obtain: $$\overline{B_Y} \subset r\cdot \overline{T(B_X)}$$ (Here $B_X$ and $B_Y$ are unit balls centered at the origin in Banach spaces $X$ and $Y$, resp. Also $r > 0$ and $T$ is linear and continuous)
Therefore, since $\overline{B_Y}$ is the closed unit ball in $Y$, for each $y\in Y$ and $\epsilon > 0$, there is an $x \in X$ for which: $$\|y - T(x)\| < \epsilon \text{ and } \|x\| \leq r \cdot \|y\|$$
Certainly, by inclusion of the balls, the first inequality is true. I don't understand where the second inequality is coming from though -- where does it come from?
Given the provided set inclusion, take $y \in Y \setminus \{0\}$. Then $\frac{y}{\|y\|} \in \overline{B_Y} \subseteq r \cdot\overline{T(B_X)}$. Therefore, there exists some $z \in \overline{T(B_X)}$ such that $\frac{y}{\|y\|} = rz$. Given $z \in \overline{T(B_X)}$, there must be some $v \in B_X$ such that $\|T(v) - z\| < \frac{\varepsilon}{r\|y\|}$. Putting this together, $$\|T(v) - z\| < \frac{\varepsilon}{r\|y\|} \implies \Big\|\|y\|rT(v) - \|y\|rz\Big\| < \varepsilon \implies \|T(r\|y\|v) - y\| < \varepsilon.$$ Take $x = r\|y\|v \in X$ such that $\|x\| = \Big\|r\|y\|v\Big\| = r\|y\| \cdot \|v\| \le r\|y\|$ as required.
If $y = 0$, then take $x = 0$.