Let $A\subset \mathbb{R}$ be a bounded set with $s = \sup A$. I want to show that there is a sequence of elements of $A$ which converges to $s$. I tried as follows.
Since $s$ is supremum of $A$ we know that for all $\epsilon > 0$ there is $x\in A$ with $s-\epsilon < x \leq s$. In that case, for each $n\in \mathbb{N}$ there is $x_n\in A$ with $s-\frac{1}{n}< x_n \leq s$. Consider the sequence $(x_n)_{n\in \mathbb{N}}$. I claim that $x_n\to s$. Before anything else, notice that if $k,j\in \mathbb{N}$ are given and $j < k$ then $1/k < 1/j$ and
$$s- \dfrac{1}{j} < s - \dfrac{1}{k} < x_k \leq s,$$
in other words, we have $x_j \in (s-1/j,s]$ and also $x_k\in (s-1/j,s]$ when $j < k$ by construction.
For that matter, suppose, for the sake of contradiction, that $(x_n)$ doesn't converge to $s$. In that case there is some $\epsilon > 0$ such that for each $n_0 \in \mathbb{N}$ there is $n\in \mathbb{N}$ with $n > n_0$ and $|x_n-s|\geq \epsilon$. Now, $(\mathbb{R},+,\cdot,\leq)$ is archimedian, in which case that for this $\epsilon$ there is $m\in \mathbb{N}$ such that $m > 1/\epsilon$ or $\epsilon > 1/m$.
From the fact that $(x_n)$ doesn't converge to $s$ we have that for each $n_0\in \mathbb{N}$ there is $n> n_0$ with $|x_n-s|\geq \epsilon > 1/m$. In that case, setting $n_0 = m$ we see that there is $N\in \mathbb{N}$ with $N > m$ and $|x_n-s|> 1/m$, in other words $x_N \notin (s-1/m,s]$ which is a contradiction, so that we must have $x_n\to s$.
Is this proof correct? I'm a little unsure on the way I've constructed the sequence. Is there any point that can be improved?
It is correct, but there is no need to use contradiction.
Let $\epsilon>0$ be given. Since $\mathbb N$ is not bounded above, there is some $N\in\mathbb N$ such that $\epsilon>1/N.$ If $n\geq N$ then $s-\epsilon<s-1/N\leq s-1/n<x_n\leq s$ which implies that $|x_n-s|<\epsilon.$ Thus, for all $n\geq N$ $|x_n-s|<\epsilon$ and hence $\lim\limits_{n\to\infty}x_n=s.$