# Isometry of a complete normed space is also complete.

Let $X$ be a complete normed space and assume the normed space $Y$ is isometric to $X$. Show that $Y$ is complete.

I tried:

Since X is complete $||x_n-x_m||<\epsilon, \forall n,m>N$ and since $Y$ is isometric to $X$ there exists an isometry $f:X\to Y$ such that $$||f(x_n)-f(x_m)||=||x_n-x_m||<\epsilon, \forall n,m>N.$$ I stuck at this step.

Pick an arbitrary Cauchy sequence $\{y_n\}\subset Y$, and let $f\colon X\to Y$ be the isometry. For each $n\geq1$, $y_n=f(x_n)$ for some $x_n\in X$. We have \begin{equation*} \|y_n-y_m\|=\|f(x_n)-f(x_m)\|=\|x_n-x_m\|, \end{equation*} so that $\{x_n\}$ is a Cauchy sequence in $X$. Since $X$ is complete, $x_n\to x$ for some $x\in X$. Therefore, $y=f(x)$ is the limit of $\{y_n\}$, since \begin{equation*} \|y_n-y\|=\|f(x_n)-f(x)\|=\|x_n-x\| \end{equation*}
Therefore, $Y$ is also complete.
Ok so let $\{y_n\}$ be a Cauchy sequence in $Y$. Write $y_n=f(x_n)$ for some $x_n\in X$. As $f$ is an isometry, $\{x_n\}$ is a Cauchy sequence in $X$ and thus has a limit $x$, say. Then it is immediate that $f(x)$ is the limit of $\{y_n\}$. (The result more generally holds for metric spaces. I also note that some people don't require isometries to be surjective, here of course you assume that they are.)
There is nothing to prove here: $Y$ is a complete metric space even if it wouldn't be a vector space. It's all in the word "isometric". This word says that $Y$ is a bijective copy of $X$ whereby the distance between points is preserved. This implies that the notions of "convergence" or "Cauchy sequence" in $X$ and in $Y$ are the same.
You've got the idea, but your proof is lacking. What is $\varepsilon$? What are $x_m$ and $x_n$?
Here's my suggestion: fix an isometry $f$ from $Y$ to $X$, let $(y_n)_{n=1}^\infty$ be a Cauchy sequence in $Y$, transport it to $X$ via $f$, and show that this new sequence is Cauchy in $X$. Since $X$ is complete, $(f(y_n))_{n=1}^\infty$ has a limit $x$ in $X$. Since $f$ is surjective, $x = f(y)$ for some $y$ in $Y$. Now show that $y_n \to y$.