I am reading a proof and it states the following without proof: Two separable Banach spaces $X$ and $Y$ are isomorphic iff there are sequences $(x_n)\subset X$ and $(y_n)\subset Y$ such that $\overline{sp}\{x_n \mid n\in\mathbb{N}\}=X$ and $\overline{sp}\{y_n \mid n\in\mathbb{N}\}=Y$ and $c\|\sum\limits_{n=0}^\infty a_nx_n\|\leq \|\sum\limits_{n=0}^\infty a_ny_n\|\leq C\|\sum\limits_{n=0}^\infty a_nx_n\|$for some $c,C>0$ and all $(a_n)\subset \mathbb{R}$.
I can prove <= easily but I am having trouble with the other direction. So far I have taken countable dense subsets $(x_n)$ of $X$ and $(y_n)$ of $Y$. These then automatically satisfy the closed linear span requirement. I am not sure how to verify the norm requirement. I know it must use the fact that $X$ and $Y$ are isomorphic but can't see how.
Could anyone please help me? Thank you
Let $\{x_n\}_{n\in \mathbb{N}}$ be a set with dense linear span and $T:X\rightarrow Y$ be an isomorphism. Put $y_n=Tx_n$. Then $\{y_n\}_{n\in \mathbb{N}}$ has dense linear span in $Y$. Now let $(a_n)$ be a sequence such that $\sum_{n=1}^\infty a_n x_n$ exists. For every $N$ the following inequality is true: $$\|\sum_{n=1}^N a_n x_n\|=\|T^{-1}\sum_{n=1}^N a_n Tx_n\|=\|T^{-1}\sum_{n=1}^N a_n y_n\|\le\|T^{-1}\|\|\sum_{n=1}^N a_n y_n\|.$$ Similarly one obtains $$\|\sum_{n=1}^N a_n y_n\|\le \|T\|\|\sum_{n=1}^N a_n x_n\|,$$ so $$\frac{1}{\|T^{-1}\|}\|\sum_{n=1}^N a_n x_n\|\le\|\sum_{n=1}^N a_n y_n\|\le \|T\|\|\sum_{n=1}^N a_n x_n\|$$ for all $N$. Taking limits yields the claim.