A set $X \in \mathbb R$ is said to be upper bounded when there is some $b \in \mathbb R$ such that $x \le b$ $\forall x \in X$. Then there is an element $y \in X$ such that $a \lt y \le b$ such that $y+ \epsilon \ge b$ does not belong to X. However, by definition, a set $X \subset \mathbb R$ is called dense in $\mathbb R$ when every open interval (a, b) contains some point of X. Therefore $X$ is not dense in $\mathbb R$.
It is clear to me that if the set is limited then there is an element within that set that when added to any value, this sum is no longer part of the set. That's what I tried to show above, I don't know if I succeeded and I don't know if my reasoning is enough to show that in fact a limited set is not dense in $\mathbb R$.
Thanks for any help.
If $b$ is an upper bound of $X$, then $(b,b+1)$ is a non-empty open set which has no element of $X$. Therefore, $X$ is not empty.
Your proof doesn't work. You don't say what $a$ is. Furthermore, you claim that $(a,b)$ contains points of $X$ (by the, this is not true in general). How could you deduce from this that $X$ is not dense?