Metric Space Completeness is Preserved by Isometry

Theorem
Let $M_1 = \left({A, d_1}\right)$ and $M_2 = \left({B, d_2}\right)$ be metric spaces.

Let $\phi: M_1 \to M_2$ be an isometry.

If $M_1$ is complete then so is $M_2$.

Proof 1
Let $\tau_1$ be the topology on $A$ induced by $d_1$ and let $\tau_2$ be the topology on $B$ induced by $d_2$.

Let $\langle x_n \rangle$ be a Cauchy sequence in $B$.

Since Inverse of Isometry is Isometry, $\phi^{-1}$ is an isometry.

Since Isometric Image of Cauchy Sequence is Cauchy Sequence, $\langle \phi^{-1}(x_n) \rangle$ is a Cauchy sequence.

Since $M_1$ is a complete metric space, $\langle \phi^{-1}(x_n) \rangle$ converges.

Since Isometry Preserves Sequence Convergence, $\langle \phi(\phi^{-1}(x_n)) \rangle$ converges.

But $\langle \phi(\phi^{-1}(x_n)) \rangle = \langle x_n \rangle$, so $\langle x_n \rangle$ converges.

Thus each Cauchy sequence in $M_2$ converges, so $M_2$ is a complete metric space.

Proof 2
Let $\epsilon >0$. Let $(b_n)$ be a Cauchy sequence in $B$. Then we have the existence of some $N_1\in \N$ such that $$d_2(b_n,b_m)< \epsilon$$ whenever $n,m\ge N_1$ and $b_n,b_m \in B$. Now since $M_1$ is isometric to $M_2$ and being isometric is an equivalence relation, and in particular symmetric, that $M_2$ is isometric to $M_1$, via $\phi^{-1}$. Thus, $$d_1(\phi^{-1}(b_n),\phi^{-1}(b_m)) = d_2(b_n,b_m)< \epsilon$$ whenever $n,m\ge N_1$ and $\phi^{-1}(b_n),\phi^{-1}(b_m) \in A$ and so $(\phi^{-1}(b_n))$ is Cauchy in $A$. Since $A$ is complete, $\phi^{-1}(b_n)$ converges in $A$ to, say, $a$. Since $\phi^{-1}$ is a bijection, and in particular surjective, there exists some $b \in B$ such that $\phi^{-1}(b) = a$. Since $\phi^{-1}(b_n)$ converges to $\phi^{-1}(b)$, there exists some $N_2\in \N$ such that $$d_1(\phi^{-1}(b_n),\phi^{-1}(b))< \epsilon$$ whenever $n\ge N_2$ and $\phi^{-1}(b_n),\phi^{-1}(b) \in A$. Now, since $M_2$ is isometric to $M_1$, we have $d_1(\phi^{-1}(b_n),\phi^{-1}(b))= d_2(b_n,b)$ and so $$d_2(b_n,b)< \epsilon$$ whenever $n\ge N_2$ and $b_n,b \in B$. Thus, $b_n$ converges in $B$ and since it was an arbitrary Cauchy sequence, we have that $M_2$ is complete, as required.