Hello everyone, I wanted to see if my proof is enough. Thank you for your critique.
Let $S$ and $T$ be nonempty sets of $\mathbb{R}$ and $S\subset\ T$, and $s$ be an arbitrary element of $S$. Our claim is that $s\leq t$ for some $t\in T$. By contradiction, let us suppose $s>t$ for all $t\in T$. Now, let $s=supS$ and $t=supT$, and so $supS>supT$. Then $supT$ is not an upper bound for B. This means that $\exists s^{'}\in S$, s.t. $t<s^{'}$. But $S\subset T$ so $s^{'}\subset T$, so $\exists s{'}\in T$, s.t. $t<s{'}$. But this is a contradiction since $t=supT$ is an upper bound of T. Hence it must be that $s\leq t$. $\blacksquare$
If $S \subset T$ and $s \in S$, then $s \in T$ and $s \leq s$ suffices.