So, you have shown that $\inf_T(S)\leq \inf_U(S)=u$. You want to show that $u\leq\inf_T(S)$. Clearly, you need to use the existence of $S'$ somehow, because without the existence of $S'$ you can trivially show this is false.
You know that $\inf_T(S)\leq\inf_U(S)$. You want to show that you do not have $\inf_T(S)\lt\inf_U(S)=u$.
Edit: (added case) If $S'=\emptyset$, then $u=\inf_U(\emptyset)=\sup(U)$, so in particular $\inf_T(S)\leq u$ and we are done. So we may assume that $S'\not= \emptyset$.
Added later: Actually, this is not needed strictly speaking, but it may be a good point to mention explicitly; if $t\lt u$, then $t$ is not an upper bound for $S'$; the only way that can occur in the first place is if $S'$ is not empty, since everything is an upper bound for the empty set, so this case is covered by the assumption being made.
If $t\in T$ is such that $t\lt u$, then since $t\in U$ and $u=\sup_U(S')$, you know that there exists $s'\in S'$ such that $t\lt s'\leq u$.
Edit: (implicitly assumed certain things were comparable, which may not be the case in general; fixed below).
Note that for every $s'\in S'$ and $s\in S$, you have $s'\leq u\leq s$ (since $u=\inf_U(S)$), and so in particular $s'\leq \inf_T(S)$, since $S'\subseteq T$. So $\inf_T(S)$ is comparable to every element of $S'$.
Suppose that $t\in T$ is such that $t\lt u$. Since $u=\sup_U(S')$, it follows that $t$ cannot be an upper bound for $S'$ in $U$. In particular, there exists $s'\in S'$ such that $s'\not\leq t$; so either $t\lt s'$, or else $t$ and $s'$ are incomparable.
In summary, for every $t\in T$, if $t\lt u$ and $t$ is comparable to every element of $S'$, then $t\lt\inf_T(S)$.
Since $\inf_T(S)\in T$, $\inf_T(S)$ is comparable to every element of $S'$ (as noted above), and $\inf_T(S)\not\lt\inf_T(S)$, you can can conclude by contrapositive that $\inf_T(S)\not\lt u$. Since you already know that $\inf_T(S)\leq u$, you are done.