Prove through approximation property

Question

Let X be a non-empty subset of $\mathbb{R}$ that is bounded above. Assume that $sup(X) \notin X$. I mus t prove that there exists an increasing sequence $(x_n)_{n=1}^\infty$ of X so that $lim_{n \rightarrow \infty} x_{n}= sup (X)$. So far I have, by the approximation property for suprema, for all $\varepsilon >0$ there exists $x_{\varepsilon} \in X$ such that $sup(X)-\varepsilon <x_{\varepsilon} \geq sup(X)$. For an increasing sequence $(x_n)_{n=1}^\infty$ of X to have $lim_{n \rightarrow \infty } x_{n}=sup(X)$, it must converge to sup(X) and by the definition of convergence $(X_n)_{n=1}^\infty$ converges to sup(X) if for all $\varepsilon >0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $|x_{n}-sup(X)|< \varepsilon$. Now don't know how to complete the proof. I am also not very confident in what I have so far, so that might be wrong.

Answer 1

By the approximation propery of suprema for $\epsilon=1,\frac{1}{2},\frac{1}{3}...\frac{1}{n}...$ we can find $x_n \in X$ such that $$\sup(X)-\frac{1}{n}< x_n \leqslant \sup(X)$$

You can use the density of rational or irrational numbers on the real line to construct an increasing sequence.

You can continue it from here.