# If $d_1(x,y) \leq d_2(x,y)$ and all $d_2$-closed balls in $M$ are also $d_1$-closed, then $(M,d_2)$ is complete.

Let $(M,d_1)$ be a complete metric space and let $d_2$ be a metric on $M$ with $d_1(x,y) \leq d_2(x,y)$ for all $x,y \in M.$ If all $d_2$-closed balls in $M$ are also $d_1$-closed, then $(M,d_2)$ is complete.

Suppose $(x_n)$ is a Cauchy sequence in $(M,d_2)$. I want to prove that the sequence converges in $(M,d_2).$

By definition of Cauchy in $(M,d_2)$, we have $d_2(x_n,x_m) < \varepsilon$ for all $m>n \geq N.$ Since $d_1(x,y) \leq d_2(x,y)$, we have $(x_n)$ is Cauchy in $(M,d_1).$ Since $(M,d_1)$ is complete, there exists $x \in (M,d_1)$ such that $x_n$ converges to $x$. Then I stuck here.

I do not know how to apply the assumption involving closed balls. Any hint would be appreciated.

By your argument so far, for all $m, n\geq N$, $x_m\in B_{d_2}(x_n,\epsilon)$. Here $B_{d_2}(x_n,\epsilon)$ stands for the closed ball centered at $x_n$ with radius $\epsilon$ with respect to the $d_2$ metric. Since such a ball is also closed in $d_1$, and $x_n\to x$ with respect to $d_1$, $x\in B_{d_2}(x_n,\epsilon)$ for all $n\geq N$. This proves $x_n\to x$ also in the $d_2$ metric.