I don't see a really slick proof, so here's a rough (and fairly standard) one. Assume that $U$ satisfies (1,2,3,4'). Notice that, by (2) and (3), $U$ is closed under subsets, so the cardinalities of elements of $U$ form an initial segment of the cardinal numbers; let $\kappa$ be the smallest cardinal not in this initial segment. It follows from (1) that $\kappa$ is uncountable, from (3) that whenever $\mu<\kappa$ then $2^\mu<\kappa$, and from (4') that $\kappa$ is a regular cardinal. So $\kappa$ is strongly inaccessible. If $U$ contained any sets of rank $\geq\kappa$, then the lowest-rank such sets would, by (2), have rank exactly $\kappa$; but this is impossible, because (by regularity of $\kappa$) any set of rank exactly $\kappa$ must have cardinality at least $\kappa$. The conclusion, therefore, is that all sets in $U$ have rank $<\kappa$. With the usual notation for the levels of the cumulative hierarchy, we have $U\subseteq V_\kappa$.
Next, notice that for all $\alpha<\kappa$, $V_\alpha$ has cardinality $<\kappa$. (This is proved by induction on $\alpha$, using at successor stages that $\mu<\kappa$ implies $2^\mu<\kappa$, and using at limit stages that $\kappa$ is regular and uncountable.) Therefore, $V_\kappa$, being the union of $\kappa$ sets $V_\alpha$ each smaller than $\kappa$, has cardinality only $\kappa$. In view of the result in the preceding paragraph, we have $|U|\leq \kappa$.
This means that, if $x$ is as in the statement (4) that we want to prove, then $|x|<\kappa$, and therefore there exists some $a\in U$ with $|a|=|x|$. Fix such an $a$ and fix a bijection $f:a\to x$. Applying (4'), we find that the union $y$ of all the members of $x$ is an element of $U$. But each member of $x$ is a subset of $y$ and therefore an element of the power set of $y$. So we have $x\subseteq\mathcal P(y)$ and, by (3), $\mathcal P(y)\in U$. We already noted above that $U$ is closed under subsets, so it follows that $x\in U$, as desired.
