Solutions

a)
i) We have $y<x<\alpha$. Hence, by definition, $y\in A$.
ii) $\alpha-1$ is in $A$.
iii) $\alpha+1\notin A$, but $\alpha+1\in\mathbb{R}$.
iv) Let

$$x'=x+\frac{\alpha-x}{2}.$$

b) We just need to prove that $\alpha=\sup A$. We have $A\ne\varnothing$. Let $a\in A$. There exists $\delta>0$ such that $a+\delta\notin A$. Otherwise, $A=\mathbb{R}$ with point i). Hence, $A$ is bounded above by $a+\delta$. This means that $\sup A$ exists. If $x<\sup A$, then there must exist $y\in A$ such that

$$x<y<\sup A.$$

Hence, $x\in A$ by i).

Now, for contradiction, suppose that $x\in A$ and $x\ge\sup A$. By iv), there exists $y\in A$ such that $x<y$. But this implies

$$\sup A\le x<y,$$

which contradicts the definition of upper bounds.