Given two Banach spaces $X$ and $Y$.
(More generally locally convex spaces)
Regard a closed subspace $U\subseteq X$.
Does every bounded operator extend: $$T\in\mathcal{B}(U,Y)\implies T_E\in\mathcal{B}(X,Y)$$
I guess not but I couldn't come up with a counterexample..
No. Say $X=L^1(\Bbb T)$, and let $$Y=U=H^1(\Bbb T)=\{f\in L^1(\Bbb T):\hat f(n)=0,n<0\}.$$So $H^1$ is just the classical Hardy space, the class of functions that extend to holomorphic functions in the unit disk.
Define $T:U\to Y$ by $Tf=f$. If $T$ extended to an operator on $X$ that would be a projection from $L^1$ onto $H^1$, and there is no such projection. (See Rudin Functional Analysis: An averaging argument shows that if there is a bounded projection then the "obvious" projection would work, but it's not bounded.)
Come to think of it, we get a similar example if $X$ is any Banach space and $U$ is a non-complemented subspace; the above is the only non-complemented subspace I happen to know. (Ok, there's also $H^\infty\subset L^\infty$.)