Show that a weak-to-norm continuous operator is compact The following screenshot is taken from the book 'Topics in Banach Space Theory'.

I don't understand how the author obtains the second sentence, which states that 
'Therefore in order to prove that $T$ is compact it suffices to show that $T|_{B_X}$ is weak-to-norm continuous'.  Here, $B_X$ refers to the closed unit ball of $X$.
Can anyone enlighten me?
 A: To show that T is a compact operator it suffices to show that the closure of the image of the closed unit ball $B_X$ is compact. Now the continuous image of a compact space is compact.And $B_X$ with the weak topology is a compact space. So if $f$ is any function $f:B_X\to Y$ which is continuous with respect to the weak topology on $B_X$,then the image $f(B_X)$ is a compact space. In this Q, with $f=T$ ,the target space is $Y=l_p$ with the $l_p$-norm topology. If $T$ is continuous with respect to these topologies then $T(B_X)$ is compact and also closed in $l_p$ (because it's compact in the metric space $l_p$.)
A: Let us suppose that $T|_{B_X}$ is weak-to-norm continuous.  To show that $T$ is compact, we must show that $T(B_X)$ is precompact in $Y$.  But $B_X$ is compact with respect to the weak topology on $X$ and $T|_{B_X}$ is continuous with respect to that topology.  Hence $T(B_X)$ is compact, being the continuous image of a compact set (using the weak topology on $B_X$).  In particular, this means $T(B_X)$ is precompact, so $T$ is compact.
