Extension of Bounded Linear Operator Let $X$ and $Y$ be normed spaces and let $W \subset X$ be a linear subspace. Suppose that $T_W \in B(W,Y)$ and that the range of $T_W$ is finite dimensional.

I know that $T_W$ can be extended to $T \in B(X,Y)$ with $\|T\|=\|T_W\|$. How about the range?
Let $V=\overline{W} $ be the closure of $W.$ Let $x_0\in V$ and let $(x_n ), (y_n ) $ be a two sequences of elements of $W$ convergent to $x_0 $ then $$||T_W (x_n ) - T_W (y_n ) || \leq ||T_W|| ||x_n -y_n ||\to 0.$$
Hence for any $x_0 \in V$ there exist exactly one $y_0\in \overline{T(W)} =T(W) $ such that foe any sequence $(x_n ) , x_n \in W$ such that $x_n \to x_0 $ the sequence $(T(x_n ))$ tends to $y_0 .$
So ve can extend $T_W $ to closed subspace $V$ by $T(x_0 ) =y_0 $ for $x_0 \in V\setminus W.$
Now from the closed graph theorem there exists a projection $P: X\to V,$ such that $P(x) =x $ for $x\in V.$ So we can define $T=T_V \circ P. $
Let $V=\overline{W} $ be the closure of $W.$ Let $x_0\in V$ and let $(x_n ), (y_n ) $ be a two sequences of elements of $W$ convergent to $x_0 $ then $$||T_W (x_n ) - T_W (y_n ) || \leq ||T_W|| ||x_n -y_n ||\to 0.$$
Hence for any $x_0 \in V$ there exist exactly one $y_0\in \overline{T(W)} =T(W) $ such that foe any sequence $(x_n ) , x_n \in W$ such that $x_n \to x_0 $ the sequence $(T(x_n ))$ tends to $y_0 .$
So ve can extend $T_W $ to closed subspace $V$ by $T(x_0 ) =y_0 $ for $x_0 \in V\setminus W.$
Now from the closed graph theorem there exists a projection $P: X\to V,$ such that $P(x) =x $ for $x\in V.$ So we can define $T=T_V \circ P. $