First, we note that all the terms $\sup(S)$, $\sup(T)$ and $\sup(S\cup T)$ exist. We now divide the proof into two parts.
Part 1: We show that $\sup(S)\le\sup(S\cup T)$ and $\sup(T)\le\sup(S\cup T)$. (Hence $\max\{\sup(S),\sup(T)\}\le\sup(S\cup T)$.)
Since $S\subset S\cup T$, we have $\sup(S\cup T)$ is an upper bound of $S$. Since $\sup(S)$ is the least upper bound of $S$, we have $\sup(S)\le\sup(S\cup T)$. Similarly, we can prove that $\sup(T)\le\sup(S\cup T)$.
Part 2: We show that $\sup(S\cup T)\le\max\{\sup(S),\sup(T)\}$.
To prove this statement, we show that $\max\{\sup(S),\sup(T)\}$ is an upper bound of $S\cup T$. Let $x\in S\cup T$. This implies that $x\in S$ or $x\in T$. If $x\in S$, then $x\le\sup S\le \max\{\sup(S),\sup(T)\}$. If $x\in T$, then $x\le\sup T\le \max\{\sup(S),\sup(T)\}$. Since $\sup(S\cup T)$ is the least upper bound of $S\cup T$, we have $\sup(S\cup T)\le\max\{\sup(S),\sup(T)\}$.
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