It's easier to imagine when we consider specific subsets.
Let's consider the subset $A=${$1, 1.4, 1.41, 1.414, 1.4142, ...$}, where the decimals approach $\sqrt{2}$.
The least upper bound of this subset, in $\mathbb{R}$, is $\sqrt{2}$. Every decimal in $A$ is less than $\sqrt{2}$, so $\sqrt{2}$ is an upper bound. And since there are decimals in $A$ that get arbitrarily close to $\sqrt{2}$, no number less than $\sqrt{2}$ can be an upper bound for $A$. Therefore, $\sqrt{2}$ is the least upper bound of $A$.
On the other hand, in $\mathbb{Q}$, there is no least upper bound. Why? Suppose there were a least upper bound $L$. Clearly $L>\sqrt{2}$. But then there's another rational $L'$ between $L$ and $\sqrt{2}$, and $L'$ is also an upper bound for $A$. So $L$ wasn't the least such upper bound after all, contradiction.
It was easy to show there that $\mathbb{Q}$ doesn't always have least upper bounds, but showing that any bounded subset in $\mathbb{R}$ does have a least upper bound is much trickier. See, e.g., a bottom-up construction of the reals by Dedekind cuts if you want a formal proof of that.