I am trying to show that:
If there are sequences $(x_n)\subset X$ and $(y_n)\subset Y$ where $X$ and $Y$ are separable Banach spaces, such that $\overline{sp}\{x_n \mid n\in\mathbb{N}\}=X$ and $\overline{sp}\{y_n \mid n\in\mathbb{N}\}=Y$ and $c\|\sum\limits_{n=1}^\N a_nx_n\|\leq \|\sum\limits_{n=1}^N a_ny_n\|\leq C\|\sum\limits_{n=1}^N a_nx_n\|$for some $c,C>0$ and all $N\in\mathbb{N}$, $(a_n)_{n=1}^N\subset \mathbb{R}$, then $X$ and $Y$ are isomorphic.
I am not sure how to proceed. The approach i think i can use is: If I define $T : sp\{x_n \mid n\in\mathbb{N}\} \to sp\{y_n \mid n\in\mathbb{N}\}$ by $T\left(\sum\limits_{n=1}^N a_nx_n\right)=\sum\limits_{n=1}^N a_ny_n$, showing this is an isomorphism, I could extend to $X$ and $Y$ by density.
Again, I am unsure if this approach works. Any help would be appreciated.
First note that the way you are defining $T$ does not give you guarantee that $T$ will be linear. To make sure that $T$ is linear you need $\{x_n \mid n \in \mathbb{N}\}$ and $\{y_n \mid n \in \mathbb{N}\}$ to be linearly independent, which you can assume without loss of generality. Because if they are not linearly independent then you can choose a maximal linearly independent set in them.
So, WLOG we can assume that $\{x_n \mid n \in \mathbb{N}\}$ and $\{y_n \mid n \in \mathbb{N}\}$ are linearly independent. Define $T:Span\{x_n \mid n \in \mathbb{N}\} \to Span \{y_n \mid n \in \mathbb{N}\}$ by \begin{equation} T\left(\sum_\limits{k=1}^{n}a_nx_n\right) = \sum_\limits{k=1}^{n}a_ny_n. \end{equation} Now use the density to extend this operator $T$ on the whole of $X$. Note that by the hypothesis there exist $c,C>0$ such that $c\|x\|\leq \|Tx\|\leq C\|x\|$ for all $x\in X$. Now observe that $\|Tx\|\geq c\|x\|$ for all $x\in X$ shows that $T$ is injective.
To see the surjectivity of $T$ notice that if $y\in Y$ is any element then choose a sequence $(y_n)$ in $Span \{y_n \mid n \in \mathbb{N}\}$ such that $y_n \longrightarrow y$. Since $T:Span\{x_n \mid n \in \mathbb{N}\} \to Span \{y_n \mid n \in \mathbb{N}\}$ is clearly surjective, so, you get a sequence $(x_n)$ in $Span\{x_n \mid n \in \mathbb{N}\}$ such that $Tx_n = y_n$. Since $(y_n)$ is convergent it is Cauchy in particular and hence by the condition $\|Tx\|\geq c\|x\|$ it follows that $(x_n)$ is Cauchy. Since $X$ is a banach space it follows that $x_n \longrightarrow x$ for some $x\in X$. Now the condition $\|Tx\|\leq C\|x\|$ shows that $T$ is continuous and hence $Tx_n \longrightarrow Tx$. But $Tx_n \longrightarrow y$, therefore, $Tx = y$. Hence $T$ is surjective.
This completes the proof and shows that $X$ and $Y$ are isomorphic.