Assume {$a_n$} is bounded sequence with the property that every convergent subsequent of {$a_n$} converges to the same limit a $\mathbb{R}$ show ... solution
Assume {$a_n$} is bounded sequence with the property that every convergent subsequent of {$a_n$} converges to the same limit a R show that {$a_n$} must converge to a. 
I don't understand that why {$a_{n_j}$} cannot converge to a...
So if the sequence does't converge to a, then its' subsequence cannot converges to a ? 
Also, in this problem 
Does "{$a_n$} does not converge to a" means that {$a_n$} is divergent ? or just {$a_n$} is convergent but not converges to a 
I am confused.    
 A: I'll not look at the provided solution, so maybe you'll get a different insight.
It's quite easy if you know about limes superior and limes inferior.
Since the given sequence is bounded, both limits (inferior and superior) are finite. There is a subsequence converging to the limes inferior and one converging to the limes superior. By assumption, these converge to the same number, therefore
$$
\liminf_{n\to\infty}a_n=\limsup_{n\to\infty}a_n
$$
and so the sequence converges.

Without the above concept, you can do as follows. First there is a converging subsequence, since the given sequence is bounded (Bolzano-Weierstraß). So we can assume, by contradiction, that the sequence doesn't converge to $a$ (the limit of all convergent subsequences).
This can mean it doesn't converge at all or that it converges to somewhere else than $a$, but it's not relevant. The important thing is that

there exists $\varepsilon>0$ such that, for all $N$, there is $n>N$ with $|a_n-a|\ge\varepsilon$.

Now choose $n_0>0$ such that $|a_{n_0}-a|\ge\varepsilon$. Next choose $n_1>n_0$ such that $|a_{n_1}-a|\ge\varepsilon$ and go on.
More precisely, if you have already chosen $n_k$ such that $n_0<n_1<\dots<n_k$ and $|a_{n_k}-a|\ge\varepsilon$, choose $n_{k+1}$ such that $|a_{n_{k+1}}-a|\ge\varepsilon$.
Thus we have a built a subsequence, but it's not necessarily convergent. However, since it is itself a bounded sequence, it has a convergent subsequence. This further subsequence cannot converge to $a$, because each of its terms $a_{n_{k_l}}$ satisfies $|a_{n_{k_l}}-a|\ge\varepsilon$. Contradiction.
A: Note that the subsequence $(a_{n_j})$ in step 3 is not just any subsequence; it's the subsequence of entries satisfying $|a_{n_j}-a|\geq\epsilon$ (we know from step 2 that there are infinitely many such entries, so we won't run out of them when constructing this subsequence).
Since each entry of $(a_{n_j})$ is at least $\epsilon$ away from $a$, this subsequence cannot converge to $a$. In fact any subsequence of $(a_{n_j})$ (a sub-subsequence of the original sequence) still has this property, so it also cannot converge to $a$.
Here "not converging to $a$" means it might not converge, or might converge to something other than $a$.
Here is an example which might clarify. Imagine $a=0$, $\epsilon=1$ and the sequence $(a_n)$ is
$$
  (a_n)=\left(1,2,\frac12,1,2,\frac13,1,2,\frac14,\ldots\right)
$$
The subsequence $(a_{n_j})$ might consist of all entries which are at least $1$ (we assumed there were infinitely many such entries):
$$
  (a_{n_j})=(1,2,1,2,1,2,\ldots)
$$
Now we can choose a convergent subsequence of this:
$$
  (a_{n_{j_k}})=(2,2,2,\ldots).
$$
Since all the entries are at least $1$, this can't converge to $a=0$. So we found a convergent subsequence which doesn't converge to $a$, giving a contradiction.
