Let $(x_n)$ be a Schauder basis of $X$ and $(y_n)$ an equivalent one to $Y$. They are supposed to be equivalent, hence for every sequence $(a_n)$ the series $\sum_{n \in \mathbb{N}} a_nx_n$ converges iff the one $\sum_{n \in \mathbb{N}} a_ny_n$ converges.
Now I want to show that the map $T:X \rightarrow Y,\sum_{n \in \mathbb{N}} a_nx_n \mapsto \sum_{n \in \mathbb{N}} a_ny_n$ is continuous. I wanted to do so by using the CGT because I don't see howwe could see that $||T||$ is finite using elementary math.
$z_n \rightarrow z$ and $Tz_n \rightarrow y$, then we need to show that $y = Tz$. But how do I show this? Is this a good idea using the CGT?
I assume below that you're working in Banach spaces.
Consider the biorthogonal functionals $l_i \in X^*$ for which $l_i(x_j) = \delta_{ij}$. These functionals have uniformly bounded norms: this follows from the Banach-Steinhaus Theorem (a.k.a. uniform boundedness principle). For more info, see the wiki.
We'll check that if $z_n \to z$ in $X$ and $T z_n \to y \in Y$, then $T z = y$ (the notation $\{x_n\}$ is overloaded).
If $z_n \to z$, if $a_m^{n}$ is the $x_m$ coefficient in the expansion of $z_n$, then $\{a_m^n\}$ is a cauchy sequence in $n$ converging to some $a_m$ (using the continuity of the functionals $\{l_i\}$, and $z = \sum_m a_m x_m$.
Similar reasoning can be applied to the $y_m$ expansions of $T z_m$, and from here you can show that the $y_m$ expansion of the limit $y$ must coicide with the $y_m$ expansion of $T z$, so that $y = T z$ holds.