Proof of the fact that the closure of a set always contains its supremum and an open set cannot contain its supremum

I would like to ask, if my proof checks out and is completely sound.

Exercise 3.2.4 from Stephen Abbot’s Understanding Analysis.
$newcommand{absval}[1]{leftlvert #1 rightrvert}$

Let $A$ be non-empty subset of $mathbf{R}$ and bounded above, so that $s = sup A$ exists. Let $bar{A} = A cup L$ be the closure of $A$.

(a) Show that $s in bar{A}$.

(b) Can an open set contain its supremum?

My Attempt.

(a) $bar{A} = A cup L$ is the closure of $A$ and contains the limits points of $A$. We proceed by contradiction. Assume that $s notin bar{A}$ and is not a limit point of $A$.

Since, $s$ is the supremum for $A$, looking at the definition of least upper bound, it must satisfy two properties: (i) $s$ is an upper bound for $A$. (ii) Given any small arbitrary, but fixed positive real $epsilon > 0$, $(s – epsilon)$ should not be an upper bound for $a$.

From (ii), it follows that, given any $epsilon > 0$, there exists $t in A$, such that $s – epsilon < t$. Thus,
begin{align*}
absval{t – s} < epsilon
end{align*}

Thus, every $epsilon$-neighbourhood of $s$, $V_epsilon(s)$ intersects $A$ in points other than $s$. So, $s$ is the limit point of $A$. Therefore, $s in bar{A}$, which contradicts our initial assumption. Hence, our initial assumption is false.

(b) An open set cannot contain its supremum. We proceed by contradiction. Let $O$ be an open set. Assume that $s in O$.

Since $O$ is an open set, for all points $x$ belonging to $O$, there exists an $epsilon$-neighbourhood $V_epsilon(x)$ that is contained in $O$. In particular, $V_epsilon(s) subseteq O$. So, if
begin{align*}
s – epsilon < t < s + epsilon
end{align*}

then $t in O$, for some $epsilon$. But, that implies, for some $epsilon > 0$, we must have
begin{align*}
s < t < s + epsilon
end{align*}

$t in O$. So, $s$ is not an upper bound for $O$. This is a contradiction. Our initial assumption must be false. $s notin O$.