Q:

Is there a systematic way to find bijective function between non-empty set $\mathbb{X}$ and Natural numbers?

I have read about the countable and uncountable sets. By the definition of the countable set I have to find a bijection between non empty set $\mathbb{X}$ and natural numbers. I want to know is there a systematic approach to find out such functions to specify whether the desired set is countable or not? 

P.S I mean for most things I encounter, how would I know if it's countable?

A:

A good rule of thumb to tell if a set is countable or not could be summed at the slogan : If you can give a "finite description" of all the elements in the set, then it's countable, otherwise it's not.
What I mean a bit more precisely, is that given a finite set $A = \{a_0, \ldots, a_n\}$ (seen as an alphabet), the set $A^{(\mathbb{N})}$ of finite sequences of elements of $A$ (seen as words) is in bijections with $\mathbb{N}$ via 
$$(a_0, \ldots a_n) \longmapsto \sum_{k = 0}^n a_k \, |A|^k $$
This is amounts to writing numbers in base $|A|$ (note that we only assumed the set $A$ above was a set of numbers for simplicity, it could be any finite set). In a nutshell, the set of finite words with a finite alphabet is countable (of course, by adding a space in your alphabet, finite sentences are also countable). Note that we even include nonsensical words here (because we allow arbitrary sequences), but that only makes the subset of words that make sense all the more countable. 
Applying this principle, we can see that:

$\mathbb{Z}$ is countable by taking the set of symbols $\{-,0,1,2,\ldots,9\}$ (by mapping $-12$ to the sequence $(-,1,2)$, $453$ to the sequence $(4,5,3)$ and so on...).
$\mathbb{Q}$ is also countable by taking the set of symbols $\{-,/ ,0,1,2,\ldots,9\}$ (by mapping $\frac{-23}{4}$ to the sequence $(-,2,3,/,4)$ and so on...). 
$\mathbb{N}^2$ is countable by taking the symbols $\{:,1,\ldots,9\}$ (mapping $(23,4)$ to $(2,3,:,4)$ and so on). And actually, we can extend that to $\mathbb{N}^3$, $\mathbb{N}^n$ and even the set of finite sequences of integers (using the same alphabet).
the set $\mathbb{Z}[X]$ of integer polynomials is countable using the alphabet $(+,-,X,^,0,1,\ldots 9)$

I'll let you check that you may also use that technique for the set of algebraic numbers (complex numbers that are roots of an integer polynomial), the set of periodic integer sequences...
Conversely, any countable set $X$ can appear as a subset of a set of finite sequences (number the elements in $X$, and write that number in base $10$ to get a sequence). So sets whose elements cannot be "finitely described" will not be countable. This gives the intuition that the set $\mathbb{N}^{\mathbb{N}}$ of all integer sequences will not be countable for example (but to actually prove it we usually use a Cantor's diagonal argument).

A:

By the definition of the countable set I have to find a bijection between non empty set X and natural numbers.

Not quite. The usual definition is:

Definition. (Usual) Let $X$ denote a set. Then $X$ is countable iff there exists an injection $\mathbb{N} \leftarrow X$.

I personally prefer a slightly different vantage point:

Definition. (Alternative) Let $X$ denote a set. Then $X$ is countable iff there exists a surjective, not-necessarily total function $f:X \leftarrow \mathbb{N}$.

The idea is that $f$ "counts off" or names the elements of $X$ using natural numbers, such that each element is named by at least one number (by surjectivity) and not every natural number is necessarily used (due to partiality).
Anyway, these conditions are equivalent, and we don't need the axiom of choice to prove it, because $\mathbb{N}$ is well-orderable even w/o any choice.
Onwards.
It is often easy to tell whether or not a set is countable, because:

Proposition. 
(0). A subset of a countable set is countable.
(1). A quotient of a countable set is countable.
(2). A finite product of countable sets is countable.
(3). A countable direct sum of countable sets is countable.

So for example, define $$D = \left\{a,b \in \mathbb{N} : \mathop{\exists}_{n \in \mathbb{N}} na=b\right\}$$
Then $D$ is countable using principles (0) and (2) above.
Beware: a countable product of countable sets usually won't be countable. For example, $$\left|\prod_{n\in \mathbb{N}} 2\right| = \beth_1,$$ where $\beth_1$ is defined as the cardinality of the powerset of $\mathbb{N}$. You can read about beth numbers at wikipedia (and I recommend you do so!).

