Is $\overline{T(\overline{B_X})} = \overline{T(B_X)}$?

Let $T$ be a continuous linear functional from Banach space $X$ to Banach space $Y$. Is it true that $\overline{T(\overline{B_X})} = \overline{T(B_X)}$? Here $B_X$ denotes the open unit ball of $X$, and the overline denote the closure operation (in respective strong topologies).

It seems obviously true by a diagonal argument, but I need a rigorous proof.

Yes. If $x_n \in \overline{B_X}$, $y_n = (1 - 1/n) x_n \in B_X$, and $\|T(x_n) - T(y_n)\| \to 0$, so if $z = \lim_{n \to \infty} T(x_n)$ then $z = \lim_{n \to \infty} T(y_n)$.

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