Let $\mathit f:X_1 \to X_2$ be continuous and surjective. With certain property of $d$, if $(X_1, d_1)$ is complete, then is $(X_2,d_2)$ complete? Let $\mathit f:X_1 \to X_2$ be continuous and surjective, and $d_1(p,q)\le d_2 \bigl(\mathit f(p),\mathit f(q)\bigl)$, $\forall p,q\in X_1$. 


*

*If $(X_1, d_1)$ is complete, then is $(X_2,d_2)$ complete?

*If $(X_2,d_2)$ is complete, then is $(X_1,d_1)$ complete?
I have proved the second question as below:

Suppose that $(X_2,d_2)$ is complete. Then for a Cauchy sequence $\bigl(\mathit f(x_i)\bigl)_{i=1}^\infty$ must converge, i.e. $\forall \epsilon \gt0$, $ \exists N\in\mathbb N$ such that $\forall n\gt m\gt N$, $d_2\bigl(f(x_n),f(x_m)\bigl)\lt\epsilon$, and $\lim_{x\to\infty}f(x_i)=f(x)$.
Since $d_1(x_n,x_m)\le d_2 \bigl(\mathit f(x_n),\mathit f(x_m)\bigl)\lt\epsilon$, $(x)_{i=1}^\infty$ is also a Cauchy sequence. 
Since $f$ is continuous, $\lim_{x\to\infty}f(x_i)=f(x)$ implies $\lim_{x\to\infty}x_i=x$. 
Therefore Cauchy sequence $(x)_{i=1}^\infty$ converges, which implies that $(X_1,d_1)$ is complete.

However firstly I do not think I showed that EVERY Cauchy sequence converges, and secondly for the question 1, I think it is not necessarily true but I can not find a proof or an counterexample either.
Could anyone help me improve the proof above and also share some hints or counterexamples for the question 1?

For question 1.
Let $(y_n)\subset X_2$ be a Cauchy sequence. As $f$ is surjective, there exists $(x_n)\subset X_1$ such that $f(x_n)=y_n$ for all $n$. Now $d_1(u,v)\leq d_2(f(u),f(v))$ for all $u,v$ implies that $(x_n)$ is a Cauchy sequence. The completeness of $(X_1,d_1)$ implies the convergence of $(x_n)$ towards $x\in X_1$. Finally, the continuity of $f$ implies that $(y_n)=(f(x_n))$ converges towards $y=f(x)\in X_2$.
