$T(d_1) = T(d_2)$ if and only if $d_1(x_0,x_n) \rightarrow 0 <=> d_2(x_0,x_n) \rightarrow 0 $ (for any $(x_n)$)

Question

I have following problem:

Let $d_1, d_2$ - metrics on $X$ Show that $T(d_1) = T(d_2)$ if and only if for any $x_0, x_1, ... \in X$ if $d_1(x_n,x_0) \rightarrow 0$ then $d_2(x_n,x_0) \rightarrow 0$ and if $d_2(x_n,x_0) \rightarrow 0$ then $d_1(x_n,x_0) \rightarrow 0$

I tried to prove that statement:

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$T(d_1) = T(d_2)$ and $d_1(x_n,x_0) \rightarrow 0$ So lets define $$B_n = B(x_o, d_1(x_0,x_n) + \frac1n) $$ where $B(x,r)$ is an open ball ($\frac1n$ may also be any other sequence that converges to $0$)

Because topologies are equal then $B_n$ are open in $T(d_2)$ (but $B_n$ may not be balls) $B_n$ are open neighbourhoods of $x_0$ and $dial(B_n) \rightarrow 0$ In metrical space there is a ball in every open neighbourhood so there is a sequence of balls $A_n$ (in $d_2$) so that $A_n \contains B_n$ and $r \rightarrow 0$.

And I am stuck at this point because I don't know hot to show that for big $n$: $x_n \in A_k$.

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The only idea I have is that closed sets in $d_1$ are the same as in $d_2$ (I think that it would implicate that $T(d_1)= T(d_2)$).

Closures of sets are equal because there is a theorem that $y_0 \in cl(O) <=>$ there exist $(y_n)$ that $y_n \rightarrow y_0$ Since closure of closed sets are equal those sets then any close set is closure of set. Is it true that it implies that closed sets in $d_1$ are closed in $d_2$?