Here $$ L = \{T(x) : x \in X \textrm{ and } \|x\| \leq 1\}. $$ While going through this proof, I don't understand the step in red, namely why $y-p \in \overline{L}$, could someone explain this?
I've figured out another possible way to infer that $y \in \overline{L}$. We have $p+y \in \overline{L}$ and $p \in \overline{L}$. Thus, there are sequences $\{x_n\},\{z_n\} \subset X$ with $Tx_n \to p+y$ and $Tz_n \to p$ with $\|x_n\| \leq 1, \|z_n\| \leq 1$. Then we have $T(x_n - z_n) = Tx_n - Tz_n \to p+y-p = y$, and also $\|x_n - z_n\| \leq 1$, therefore $y \in \overline{L}$, is this correct?
Collection of comments. But first, let me prominently display the definition of $L$ to reduce possible confusion
That is, $L$ is the image under $T$ of the closed unit ball in $X$.
How the author deduces that $y-p\in \overline L$.
This is because balls $B$ around 0 are symmetric ($B=-B$), so if $y \in B_{0,Y}(t)$, then $-y\in B_{0,Y}(t)$. By the same reasoning as the line prior,
$$ p-y \in \overline L.$$ Since $\overline L = \overline {-L} = - \overline L$ (this computation was carried out in the comments), $-p+y\in \overline L$, which is the claimed result.
Is your alternative proof correct?
Not as it stands; the reason being that $\| x_n \|\le 1, \|z_n \|\le 1$ does not imply $\|x_n - z_n\| \le 1$. Indeed the triangle inequality only gives $ \|x_n - z_n\| \le \|x_n\| + \|z_n\|\le 2$ which could be satisfied with equality without further restrictions on $x_n , z_n$. So something like the symmetry argument above seems necessary, though I do not know enough proofs of this theorem to make a definitive statement.