Let ${\{X_m}\}$ be a sequence in $\mathbb{R}$ that is bounded above and $A=sup{\{X_m}\}$.
I was first required to find a convergent sequence that's limit is not $A$, and I think I've done so with the sequence ${\{1,1/2,1/3,...}\}$ which has a supremum of $1$ but converges to $0$.
Next, I'm required to prove that if none of the terms of ${\{X_m}\}$ is equal to $A$, then ${\{X_m}\}$ must have a monotonic increasing subsequence that converges to $A$, and I'm entirely unsure of where to begin for this proof. Any help is appreciated!
Try to use the definition of the supremum. That $A$ is the supremum of the $X_m$ means that for every $\varepsilon>0$ there is an $k\in\mathbb{N}$ such that $$0<A-X_k<\varepsilon$$ The LHS inequality is because none of the $X_m$ is equal to $A$. First you can find a term $X_{k(1)}$ of the sequence such that $$0<A-X_{k(1)}<1$$ Then you can find a term $X_{k_2}$ such that $$0<A-X_{k_2}<\min\left(A-X_{k_1}\,,\,\dfrac{1}{2}\right)$$ using the definition of $A$ as a supremum so that $X_{k_2}>X_{k_1}$. Try to follow this idea and build a subsequence $X_{k_m}$ by induction which converges increasingly to $A$.