A very widely stated result:
A sequence $x_n \to x$ iff every subsequence $x_{n^{\prime}}$ of $x_n$ contains a further subsequence $x_{n^{\prime\prime}}$ such that $x_{n^{\prime\prime}}\to x$.
My question is that there is no need to pass on to sub-sub sequence. The statement is true for sub sequence replacing sub-sub sequence. Then why is this statement useful?? Is it the case that in several places its difficult to find a convergent sub-sequence but easy to get a convergent sub-sub sequence??
It is a useful criterion. One may not use it every day, but it comes up not too rarely.
The typical use-case that I come across, is that one has a sequence of which it is known that every subsequence has a convergent subsequence - a sequence in a sequentially compact space; for example a normal family, or the closed unit ball of a reflexive Banach space in the weak topology - and one can show that all convergent subsequences must converge to the same limit. Then the criterion gives the convergence of the full sequence.
Example:
Suppose you have a connected open $\Omega \subset \mathbb{C}$, and a holomorphic $f \colon \Omega \to \Omega$ that has an attractive fixed point $z_0 \in \Omega$, i.e. $f(z_0) = z_0$ and $\lvert f'(z_0)\rvert < 1$.
It is then clear that the sequence $f^n$ (exponent means composition here) converges uniformly to the constant function $c_{z_0} \colon z \mapsto z_0$ in a neighbourhood $U$ of $z_0$. But we don't know a priori how the sequence $\bigl(f^n\bigr)_{n\in\mathbb{N}}$ behaves on all of $\Omega$.
By the identity theorem, if a subsequence of $\bigl(f^n\bigr)_{n\in\mathbb{N}}$ converges locally uniformly, its limit must be the constant $c_{z_0}$.
Now, if $\Omega$ is nice (bounded, a half plane; $\mathbb{C}\setminus\Omega$ contains at least two points, but this one is deep), the family $\mathscr{F} = \lbrace f^n : n \in \mathbb{N}\rbrace$ is a normal family, that is, every sequence in $\mathscr{F}$ has a locally uniformly convergent subsequence.
In that case, pick any subsequence $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$ of $\bigl(f^n\bigr)_{n\in\mathbb{N}}$. Since that is a sequence in the normal family $\mathscr{F}$, it has a locally uniformly convergent subsequence $\bigl(f^{n_{k_m}}\bigr)_{m\in\mathbb{N}}$. By the identity theorem, as mentioned above, $f^{n_{k_m}}$ can only converge to $c_{z_0}$. So
By the criterion, $\bigl(f^n\bigr)_{n\in\mathbb{N}}$ itself converges locally uniformly to $c_{z_0}$.
Note that we need to pass to sub-subsequences. If we directly apply normality to $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$, all we get is that $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$ has (at least) one subsequence converging locally uniformly to $c_{z_0}$. That wouldn't rule out the possibility that $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$ has subsequences that don't converge locally uniformly.
Another use of the criterion is here.
As mentioned in some comments, the statement should be
$$x_n\xrightarrow[n\to\infty]{}x\iff \;x_{n_k}\xrightarrow[k\to\infty]{}x\;,\;\;\text{for any (every) subsequence}\;\{x_{n_k}\}\subset\{x_n\}$$
And I think you're right: the statement about the sub-subsequence is way too messy to be very helpful, so I'd rather use the above one.
The main application of the above is, imo, to prove that a sequence does not converge: it is enough to produce two subsequences that converge to different things (or that either does not even converge at all...)
