There exists strictly increasing $\{x_n\}$ that converges to $\sup E$ I need to prove that, if $E \subseteq \mathbb{R}$ is a non-empty bounded set and $\sup E \not\in E$ then there exists a strictly increasing sequence $\{x_n\}$ that converges to $\sup E$ such that $x_n \in E$ for all $n \in \mathbb{N}$.

I've been trying to find a clue in the textbook, but couldn't. I don't even know how to start the proof. Could someone please give a clue?
Hint 

Firstly, choose any element $x_1 \in E$. As stated, $\sup E \notin E \implies x_1 &lt; \sup E$. Using the approximation property for suprema, for each $n \geq 2 \exists x_n \in E$ such that $\max(\sup E -\frac{1}{n},x_{n-1})&lt;x_n&lt;\sup E$ so that a sequence $x_1&lt;x_2&lt;x_3&lt;...$ is strictly increasing.

Then go on to apply the Squeeze theorem and you should be home and dry.

Edit Show that the sequence above is increasing 

Start at $n=1$ there exists an $x \in E$ such that $\sup  E  − 1 &lt; x$. Denote this $x = x_1$. Now, for $n=2$ there exists an $x \in E$ such that $\sup E − \frac{1}{2}  &lt; x_2$. Clearly $x_2 &gt; x_1$. This works for any $n$ so there exists a sequence $x_n \in E$ with the property that $\sup E - \frac{1}{n}&lt;x_{n}$
I would phrase a hint as this:

If there were no sequence approaching $\sup E$, how could $\sup E$ be the supremum?