Let $X$ be a Banach space and $T$ a linear bounded operator defined on $L(X,Y)$ with $Y$ a normed space. If $T$ is an isometry then $TX$ is a closed subspace of $Y$.
I considered a sequence $y_n$ in $TX$. I want to show that the limit of this sequence is in $TX$. $y_n \in TX$ it means it exists $x_n \in X$ such that $Tx_n=y_n$. Being that $T$ is an isometry it is necessary a one-one map so it exists the inverse operator $T^{-1}$. So $x_n=T^{-1}y_n$. Making the limit $x=T^{-1}y$ but $x_n$ tends to $x$ which belongs to $X$ because $X$ is a Banach space. So $y=Tx \in TX$ and the thesis follows. Is my proof correct?
As PhoemueX said, the proof lacks the consideration of the continuity of $T^{-1}$. Once that is added, the proof is correct.
But I would put the proof differently:
The image of a complete metric space under a distance-preserving map is also a complete metric space.
If a subset of a metric space is complete in the induced metric, then that subset is closed.
My point here is that linearity has little to do with the problem. It's mostly the metric.