Let x_n be a bounded sequence and let x=sup {x_n: n is a natural number} and let x_n < x for all n. Prove there is a subsequence convergent to x.
First shoe that $\displaystyle \forall M \in \mathbb{Z}^+ \, \exists k \in (x_n) $ such that $\displaystyle k \in (x-\frac{1}{N},x)$. Denote this $\displaystyle k$ as $\displaystyle x_{n_M}$.
now show that the subsequence $\displaystyle (x_{n_M})$ converges to $\displaystyle x.$
You should have noticed that the key to the condition "x_n< x for all n" is that no member of the sequence is equal to x. For example, if {x_n}= {1/n} then x= sup {1/n}= 1 but {1/n} is a sequence converging to 0 so has no subsequence converging to 1. Now, the fact, that x is a sup but NOT a maximum should lead you to what abhishekkgp did.