I understand how to prove this using the approximation property of the supremum and invoking the squeeze theorem of limits. However, upon reading answers for similar questions, I think there's a more direct way of proving this without using the squeeze theorem. I'm having trouble understanding the logic though. I was wondering if I could receive some guidance. Here is what I have so far:
Since $x$ is the least upper bound of $S$, for all $n \in \mathbb{N}$ we can pick an $x_n \in S$ such that $x_n > x - \frac{1}{n}$. Since $x$ is an upper bound of $S$, $x \geq x_n$ for all $n \in \mathbb{N}$. Therefore we have $x \geq x_n > x-\frac{1}{n}$ for all $n \in \mathbb{N}$.
How do I go from here to showing that $(x_n)$ converges to $x$ using the definition of convergence (without using squeeze theorem)?
For every fixed $\epsilon >0$, there is an $N\in\mathbb{N}$ such that $\frac{1}{n}<\epsilon$ for all $n\geq N$ (this follows by the Archimedean property). Thus, for all $n\geq N$, $|x_{n}-x|=x-x_{n}<x-(x-\frac{1}{n})=\frac{1}{n}<\epsilon$.