Topologies not induced by a metric Assume that I have a space, a metric, and a possible topology on that space. The topology induced by that metric is not the possible topology.
However, we know: that, if a sequence converges in the metric, it converges in the possible topology (hypothesis). This gives us that the possible topology is finer. This is the first fact (conclusion).
We also know that, for every sequence that converges in the possible topology, one can extract a subsequence that converges in the metric. This is the second fact (hypothesis)
What additional information (conclusions) does this second fact give you? Is it that relatively compact sets in the possible topology are relatively compact in the metric, or can we say more than this? This last fact is because, given a sequence in a relatively compact in the possible topology, we know we can extract a convergent subsequence by relative compactness, and we can extract a convergent subsubsequence by the property I described, hence we have that the set is relatively compact in the metric.
I have had a colleague say that “the closures of the topologies coincide”, this did not make any sense to me because I lack a definition of what the closure of a topology is.
 A: Your first fact is false.
Let $X$ be any set. Let $\tau_d$ be a metric topology and $\tau$ be any topological space.
Let for every sequence $(x_n) \to x$ in $(X, \tau_d) $ implies $(x_n)\to x$ in $(X, \tau) $.Then $Id:(X, \tau_d) \to (X, \tau) $ is sequentially continuous. Since $(X, \tau_d) $ is first countable, $Id:X\to X$ is topological continuous . Hence $\tau\subset \tau_d$
Hence metric topology $\tau_d$ is finer than than the topology $\tau$. Infact we can choose $\tau, \tau_d$ such that $\tau$ is strictly finer than $\tau$ .
Example : Let $X$ be any uncountable set.
$\tau_d :=\tau_{\text{dis}}=\mathcal{P}(X)$
$\begin{align}\tau&:=\tau_{\text{coc}}\\&
=\{U\subset X : U=\emptyset \text { or} |X\setminus U|\le \aleph_{0}\}\end{align}$
Then $(X, \tau_{\text{dis}}) $ and $(X, \tau_{\text{coc}}) $ has same collection of convergent sequence ( all eventually constant sequences). Hence it satisfy  your given condition. However $\tau_{\text{coc}}\subset\tau_{\text{dis}}$ strictly finer as only proper closed sets in $(X, \tau_{\text{coc}})$ are countable sets.

The above example is also satisfy second condition . But contradict your "colleague's" conclusion as any proper subset of $X$  in the space $(X,\tau_{\text{coc}})$ either closed ( in case of countable sets) or dense (in case of uncountable sets) . In other words $\forall A\subset X$ , $\operatorname{Cl}_{\tau_{coc}}(A)= A\text{ or} X$.But in $(X, \tau_{dis}) $ only dense subset is $X$ itself.
Let, $A\subset X$ be any proper uncountable set. Then $\operatorname{Cl}_{\tau_{coc}}(A)=  X$ but $\operatorname{Cl}_{\tau_{dis}}(A)= A$
Hence " closures of the topologies coincide” is false.
