Any two sets of continuum cardinality are equinumerous? Is it true that any two uncountable subsets of $\mathbb{R}$  is equinumerous? If yes how to prove that? More generally can I say two sets of same Cardinality are equinumerous?
equinumerous means there is a bijective map between them
 A: You've asked two questions; I'll answer them separately.
Question 1: if two sets are of the same cardinality, are they equinumerous?
The answer is yes, by definition. However, you ask in the comments about how to prove this, so I'll elaborate more on the definitions. It's been pointed out in the comments that two sets are equinumerous if there is a bijection between them. When we say two sets have the same cardinality, we are using one of two definitions.
Usually, when we say two sets have the same cardinality, we are talking about cardinal numbers, a special kind of ordinal number (which you can look up on wikipedia). If we assume the axiom of choice, then every set $A$ is in bijection with a unique cardinal number, denoted $|A|$, which we call its cardinality. Examples include $\aleph_0$, the cardinality of countable sets such as $\mathbb{N}$ or $\mathbb{Z}$, and $2^{\aleph_0}$, the cardinality of $\mathbb{R}$. We can then say things like "$\mathbb{N}$ and $\mathbb{Z}$ have the same cardinality", because $|\mathbb{N}| = \aleph_0 = |\mathbb{Z}|$.
However, one might also just say that two sets "have the same cardinality" if they are equinumerous; that is, if there is a bijection between them. This is a definition more commonly used in beginner math courses, when you don't want to get into the details of cardinal numbers or the axiom of choice. (This second definition follows almost immediately from the first: if $A$ and $B$ both have cardinality $\alpha$, there are bijections $f : A \rightarrow \alpha$ and $g : B \rightarrow \alpha$, and so $g^{-1} \circ f : A \rightarrow B$ is a bijection too, meaning $A,B$ are equinumerous.)
Hopefully, this convinces you that "$A$ and $B$ have the same cardinality" implies "$A$ and $B$ are equinumerous", if not by definition, then almost immediately.
Question 2: are any two uncountable subsets of $\mathbb{R}$ equinumerous?
The answer to this is: it depends on the continuum hypothesis. The continuum hypothesis states that $2^{\aleph_0}$ (the cardinality of $\mathbb{R}$) is the next biggest infinity after $\aleph_0$ - there are no other sizes of infinity in between. (This is often written $\aleph_1 = 2^{\aleph_0}$.) Unfortunately, the continuum hypothesis has been shown to be independent of the ZFC axioms, so your question has no definitive answer; but, we can still consider the two cases.
If the continuum hypothesis is true, then the answer to your question is yes. Indeed, if a subset $A$ of $\mathbb{R}$ is uncountable, then we have $\aleph_0 < |A| \leq |\mathbb{R}| = 2^{\aleph_0}$, and the continuum hypothesis then implies that we must have $|A| = 2^{\aleph_0}$ ($= |\mathbb{R}|$). Thus $|A|$ must be equinumerous to $|\mathbb{R}|$. On the other hand, if the continuum hypothesis is false, then we could find an uncountable subset of $\mathbb{R}$ whose cardinality is smaller than the cardinality of $\mathbb{R}$, so the answer to your question would be no.
(There's a slight issue here because we're assuming $A$ does have a cardinality. Without the axiom of choice, I think the continuum hypothesis could be true while at the same time there exists a subset of $\mathbb{R}$ which doesn't have a cardinality, hence is not equinumerous to $\mathbb{R}$. But that's a bit beyond me; usually when talking about the continuum hypothesis, the axiom of choice is assumed.)
A: Your first question is equivalent to the continuum hypothesis (CH).
The cardinality of $\mathbb{N}$ is denoted by $\aleph_0$, and the next cardinality is denoted by $\aleph_1$. Also, it can be proven that the cardinality of $\mathbb{R}$ is equal to $2^{\aleph_0}$. CH is the hypothesis that $\aleph_1 = 2^{\aleph_0}$.
If CH is true, then there is no cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$, so any uncountable subset of $\mathbb{R}$ must have cardinality equal to $2^{\aleph_0}$.
If CH is false, then there exists an $A \subseteq \mathbb{R}$ with $|A| = \aleph_1$, but $|A| \neq |\mathbb{R}|$ of course.

The answer to your second question is that if two sets are equinumerous, i.e. there exists a bijection between them, their cardinalities are equal by definition.
