Finite dimensional subspaces Let $X$ be a complex Banach space of infinite dimension. Does there exist a finite dimensional subspace of $X$ of arbitrary (finite) dimension which is complemented by a projection of norm 1?
I am not entirely sure what you mean by "which is complemented by a projection of norm 1", but I suppose you mean: If $Y$ is a subspace of $X$, is there a non-trivial projection $P$ such that $Y\oplus P(X) = X$. If this is not what you mean, then I am sorryt for the misunderstanding. 

There exists such a projection if , and only if, the is a projection $\hat P$ such that $\hat P (X) = Y$, because then $P$ can be chosen as $1-\hat P$.

To find out for which subspaces $Y$ this is true, take the map $x \mapsto y$, where $y$ is the unique vector out of $Y$ such that $x$ can be written as $x=y+u$, for some $u$ in a complementary subspace. This map is a projection if and only if it is continuous, but now it is easy to see that this is true if and only if $Y$ is a closed subspace. 

Hence for all finite dimensional subspaces and for all infinite dim. subspaces which are closed the answer to your question is yes, otherwise no.
I am not entirely sure what you mean by "which is complemented by a projection of norm 1", but I suppose you mean: If $Y$ is a subspace of $X$, is there a non-trivial projection $P$ such that $Y\oplus P(X) = X$. If this is not what you mean, then I am sorryt for the misunderstanding. 

There exists such a projection if , and only if, the is a projection $\hat P$ such that $\hat P (X) = Y$, because then $P$ can be chosen as $1-\hat P$.

To find out for which subspaces $Y$ this is true, take the map $x \mapsto y$, where $y$ is the unique vector out of $Y$ such that $x$ can be written as $x=y+u$, for some $u$ in a complementary subspace. This map is a projection if and only if it is continuous, but now it is easy to see that this is true if and only if $Y$ is a closed subspace. 

Hence for all finite dimensional subspaces and for all infinite dim. subspaces which are closed the answer to your question is yes, otherwise no.