In $\mathbb R$,every bounded infinite set $A\subset \mathbb R$,has a limit point in $\mathbb R$.But if we replace $\mathbb R$,by any other metric space $X$,then this may not hold.For example $X=\mathbb R-\{0\}$,then $A=\{1/n:n\in \mathbb N\}$ is a bounded infinite set in $X$ but it has no limit point in $X$.But we can have a counterpart of Bolzano Weierstrass theorem in case of arbitrary metric space $X$.
First we try by replacing boundedness by total boundedness.But it fails by the same counterexample.$A$ is totally bounded in $X=\mathbb R-\{0\}$ but it has no limit point in $X$.
Now we try with complete metric spaces,
Suppose $X$ is complete,then any bounded infinite set has a limit point in $X$.
But it fails because an infinite set $X$,with discrete metric is complete and any set in $X$ is bounded,even if we take any infinite bounded set in $X$,it does not have a limit point.
So,we make the statement:
In a complete metric space,any totally bounded infinite set has a limit point in $X$.
Proof: $A$ be a totally bounded infinite set.If $A$ has no limit point,then $A$ is closed.Since $X$ is complete,so $A$ is complete.So,$A$ is complete and totally bounded and hence is compact.$A$ is infinite set in compact space $A$,so it has a limit point in $A$ and hence in $X$.
Thus we can generalize Bolzano -Weierstrass theorem.Is my thought process correct?Is there anything that I can add to this?
In a totally bounded subset $A$ (in $(X,d)$), every sequence has a Cauchy subsequence. And if $(X,d)$ is complete, this is a convergent subsequence and it follows that an infinite subset of $A$ must have a limit point in $X$.
So indeed your last attempt holds true, but I give a different reasoning. Your reasoning works too, but needs more.