Bolzano Weierstrass Theorem for General Metric Spaces Though $\mathbb{R}$ is not compact, because of LUB axiom one can conclude BW theorem i.e (every bounded sequence will have a convergent subsequence.)
My questions are:-

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*In what kind of Metric Spaces this result will hold?
(example: In Compact Metric Spaces this holds.)
Is there a charaterization to these spaces ?


*Can we find an example of a Unbounded Metric Space(like $\mathbb{R}$) where BW property holds?


*In what kind of Metric Space it is true that Every bounded sequence will have a Cauchy subsequence. (example: In Totally Bounded Metric Space this holds.) Is there a characterization to these spaces ?
 A: There are two things going on here.

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*A metric space in which every Cauchy sequence converges is called complete.


*A metric space in which every sequence has a Cauchy subsequence is equivalent to the space being totally bounded.
So what you need is a complete metric space in which every bounded subset is totally bounded.
Examples are: any compact metric space, any Euclidean space $\mathbb{R}^n$, any closed subspace of a Euclidean space.
In general this property is called the Heine-Borel property.
A: The category of metric spaces you are looking for are those in which every closed bounded subset is compact (this property is called the Heine-Borel property).
You know that if a metric space, $(X,d)$, is compact, then every sequence in $X$ has a convergent subsequence (this is known as sequential compactness). The converse is also true, by which I mean if a metric space is sequentially compact, then it is also compact.
First, suppose $(X,d)$ is a metric space in which every closed bounded subspace is compact. Let $(a_n)_{n\in\mathbb N}$ be a bounded sequence in $X$. Because this sequence is bounded, we have for some $\epsilon>0$ that $d(a_1,a_i)<\epsilon\;\forall\; i$. This means $\{a_n\}_{n\in\mathbb N}\subset B(a_1;\epsilon)\subset \overline{B(a_1;\epsilon)}$.
$Y=\overline{B(a_1;\epsilon)}$ is clearly a closed subset of $X$. It is also clearly bounded (all points are at most a distance $2\epsilon$ apart by the triangle inequality). So $Y$ is a compact subset of $X$. As a subspace of $X$, $Y$ is also a metric space. So in its own right, $Y$ is a compact metric space. Since all points of the sequence $(a_n)_{n\in\mathbb N}$ lie in $Y$, there is some subsequence $(a_{k_n})_{n\in\mathbb N}$ which converges in $Y$. As $Y\subset X$, the subsequence $(a_{k_n})_{n\in\mathbb N}$ converges in $X$ as well, so $(a_n)_{n\in\mathbb N}$ has a convergent subsequence in $X$, as desired.
For the other direction, suppose $(X,d)$ is a metric space in which every bounded sequence has a convergent subsequence. Let $A$ be any closed and bounded subset of $X$. We note immediately that $A$, being a subspace of $X$, can be treated as a metric space in its own right. Let $(a_n)_{n\in\mathbb N}$ be a sequence of points in $A$. As $A$ is a bounded subset of $X$, $(a_n)_{n\in\mathbb N}$ is a bounded sequence of $X$. By assumption then, there is some subsequence, $(a_{k_n})_{n\in\mathbb N}$, converging to some point $z\in X$. Since all elements of $(a_{k_n})_{n\in\mathbb N}$ lie in $A$ and $A$ is closed in $X$, it follows that $z$ belongs to $A$. So $(a_{k_n})_{n\in\mathbb N}$ is a convergent subsequence of $(a_n)_{n\in\mathbb N}$ in $A$, i.e. every sequence in $A$ has a convergent subsequence in $A$. This makes $A$ sequentially compact, and hence compact. As $A$ was an arbitrary closed bounded subset of $X$, all closed bounded subsets of $X$ are compact.
By the Heine-Borel theorem, all metric spaces $(\mathbb R^n,d)$ possess this Heine-Borel property (where $d$ can be either the standard Euclidean metric or the Chebyshev distance function). The metric spaces that satisfy the requirements of your second question are then exactly like euclidean space, in that they'll be noncompact spaces with the Heine-Borel property. It is also true that any noncompact space $(X,d)$ has at least one unbounded metric $d'$ that preserves its topology, but this doesn't guarantee that if $(X,d)$ has the Heine-Borel property, then $(X,d')$ does too (because different metrics create different bounded sets).
Regarding your third question, since in a metric spaces with the Heine-Borel property all bounded sequences have convergent subsequences, all bounded sequences also have Cauchy subsequences (because any convergent subsequence is trivially Cauchy). So all metric spaces with the Heine-Borel property satisfy the requirements of your 3rd question. Whether these are the only spaces that do, I don't know. It seems unlikely, but I cannot think of a good counterexample.
EDIT (due to OP/ Saikat Goswami's comment): Yes, many more spaces than just those with the Heine-Borel property satisfy the requirements of the third question. Working off OP's comment under this answer, more generally any subspace of a Heine-Borel space will satisfy the requirements of a third question (like the example of $(0,1)$ as a subspace of $(\mathbb R,d)$ where $d$ is the standard metric on the reals). The complete characterization of the spaces satisfying the third question is by the criteria in Jacob FG's answer: they are exactly the metric spaces in which all bounded sets are totally bounded.
