Subspaces of $\ell^p$ 
Is every closed subspace of $\ell^p(\Bbb N)$ isomorphic to an $\ell^p(X)$?

For $p=2$ this is true, as closed subspaces are still seperable Hilbert spaces, which have ONBs. For $p=\infty$ one has that $c_0(\Bbb N)\subset \ell^\infty(\Bbb N)$ is not isomorphic to any $\ell^\infty(X)$. Further counter-examples elude me.
I am also wondering about if the statement is true for finite dimensional subspaces when one demands isometric isomorphisms, here it is again true for $p=2$.
 A: No, this is not at all how Banach spaces work. Taking a cross-section of an irregularly shaped convex set (unit ball of $\ell^p$ for $p\ne 2$) one should expect to get very different results depending on what cross-section was taken. 
For every $p\ne 2$, the space $\ell^p$ contains a subspace that does not have a basis, and therefore is not isomorphic to $\ell^p$. Moreover, it contains a subspace that does not have approximation property (weaker than having Schauder's basis). 
That said, these are not the sort of examples that one just sits down and defines explicitly. There's some theory to be learned first. A classical reference is the book Lindenstrauss, J.; Tzafriri, L.: Classical Banach Spaces I, Sequence spaces, 1977.

for finite dimensional subspaces when one demands isometric isomorphisms

This is easier to answer; take three-dimensional cube (a three-dimensional version of the unit ball of $\ell^\infty$) and cut it by a plane. You'll get different shapes depending on the position of the plane: sometimes a square, sometimes a hexagon. These are not isometrically isomorphic. Same goes for other $\ell^p$ $(p\ne 2)$ where unit balls are rounded cubes or octahedra.
