Schauder basis under isometry.

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I am self studying linear functional analysis and am a bit confused about the following problem. It is situated after a chapter on the open mapping and closed graph theorem and in my answer I don't get close to using those, so I suspect that I am wrong.

We consider two normed spaces $X$ and $Y$ and assume that they are isomorphic. We want to show that if $X$ has schauder basis $\{e_k\}_{k=1}^{\infty}$, then $Y$ has schauder basis $\{Te_k\}_{k=1}^{\infty}$, where $T$ is an isometric mapping from $X$ into $Y$.

So the assumption goes that for each $x\in X$, there is a uniqe sequence of scalars $\{\alpha_k\}$ so that.

$$\left|\left|x-\sum_k^na_k e_k\right|\right|\to0\text{ for }n\to \infty$$

So as $T$ is bijective, we can associate each $x\in X$ with a unique $y\in Y,y=Tx$. By the preservation of distance and linearity then.

$$\left|\left|y-\sum_k^n\alpha_kTe_k\right|\right|\to 0\text{ for }\to \infty$$

Am i missing something huge here? Thanks for looking.