Let $A$ be a bounded subset of $\mathbb{R}$.
1. Show that there exists a sequence $a_n$ of elements of $A$ such that $\lim _{ }\left(a_n\right)\:=\:sup\left(A\right)$
2. Show that we can build a sequence that will be monotonic increasing that converge to $sup\left(A\right)$.
For the first question i think i need to use induction. for $a_1$ i'll choose the first element that implieas $a_1\:\ge \:supA\:-1$. then for every n, it will be : choose the $a\in A$, such that $a\:\ge \:supA\:-\:\left(\frac{1}{n}\right)$ and let $a_n\:=\:a$.
This is good enough? For the second question i don't know how to even start! tnx!
The question is trivial if $A$ is finite, so let's assume $A$ is infinite.
Since $\operatorname{sup}(A)$ is a limit point of $A$, then in particular, we can find an infinite number of points within any given $\varepsilon$-neighborhood of it. Further, all such points will be less than or equal to $\operatorname{sup}(A)$ by definition of the supremum.
So you can easily get your increasing sequence by selecting any $a_1$ and then choose $a_2$ to be such that $a_1 \leq a_2 \leq \operatorname{Sup}(A)$. Then we can find another point $a_3$ between $a_2$ and $\operatorname{Sup}(A)$. This process must be guaranteed to continue ad infinitum, else $\operatorname{Sup}(A)$ would not be a limit point of $A$. In other words, if this were not guaranteed, then you would be contradicting the very definition of the supremum of a set.
For question 1, your solution is worded a little weird. Since there are infinitely many, you won't be choosing "the" $a \in A$ such that $a \geq \operatorname{Sup}(A) - \left( \frac{1}{n} \right)$, but simply any $a$ that satisfies that condition. The right idea is there though.