This is another exercise from Allan's book "Introduction to Banach Spaces and Algebras".
Exercise 2.9: A Banach space $E$ is said to be homeomorphic to its square if $E\oplus E$ is linearly homeomorphic to $E$. Prove that the spaces $c$ and $C([0,1])$ have this property.
Note that we haven't developed much theory yet (in this book!) Well, $c_0 \oplus c_0 \cong c_0$ (split a sequence into the parts supported on the evens and odds say) and $c\cong c_0$, so that does the case of $c$. I cannot see an elementary proof for $C([0,1])$. Here's a less than easy proof: clearly $C([0,1]) \oplus_\infty C([0,1]) = C([0,1] \cup [2,3])$ say. Then appeal to Miljutin's Theorem, as $[0,1] \cup [2,3]$ is an uncountable compact metric space, $C([0,1] \cup [2,3]) \cong C([0,1])$.
Surely I don't need the full power of Miljutin's Theorem. But, for example, $C([0,1] \cup [2,3])$ cannot be isometric to $C([0,1])$ (if we use real scalars) by the Banach-Stone theorem, as $[0,1] \cup [2,3]$ is not homeomorphic to $[0,1]$.
Surely using Miljutin's Theorem is not what the book intends.
Is there an elementary proof that $C([0,1]) \oplus C([0,1]) \cong C([0,1])$?