Suppose that $S$ is a nonempty subset of $\mathbb{R}$ and $S$ is bounded above. Also suppose that $T=\{x\in\mathbb{R}:x$ is an upper bound of $S\}$ Show that $\sup(S)=\inf(T)$.
My proof: Since $S$ is nonempty bounded above subset of $\mathbb{R}$, $\sup(S)$ exists. Let $\sup(S)=\alpha$. It is clear that $\alpha\in T$ because $\alpha$ is an upper bound of $S$. Also, $\alpha\leq x$ for all $x\in T$ because $\alpha$ is a supremum. Since $\alpha\leq x$ and $\alpha\in T$, we can conclude that $\inf(T)=\alpha=\sup(S)$ as required.
Is this proof ok?
Your work is fine, rephrasing a bit:
$S \subset \mathbb{R}$, $S$ nonempty, is bounded above : $\sup(S)$ exists.
$T \subset \mathbb{R}$, $T$ nonempty (why?), is bounded below by $s \in S$: $\inf(T)$ exists.
$\sup(S)$ is the least upper bound of $S$, i.e. an upper bound, hence $\sup(S) \in T.$
Assume there is an $x \in T$ with $x \lt \sup(S)$.
Since $x$ is an upper bound of $S$ : contradiction, $\sup(S)$ is the least upper bound of $S$.
Hence for all $x \in T$: $x \ge \sup(S)$,
and since $\sup(S) \in T$, we have
$\inf(T)=\sup(S).$