If $S_j\subset \mathbb{R}^n$ and for all $j\in \mathbb{N}$ the set $S_j$ is measurable, show that $\bigcup S_j$ and $\bigcap S_j$ are measurable.
I think I can use Zorn's lemma to show that this is true.
If we let $\mathcal{M}$ be the set of all measurable subsets of $\mathbb{R}$ then we can impose a partial ordering, $\prec$, on it by set inclusion. That is $A\prec B$ iff $A\subset B$. If we define $T_i = \bigcup_{k=1}^i S_j$ then $T_i \in\mathcal{M}$ since finite unions of measurable sets are measurable.
Moreover we know that $\mathbb{R}^n$ is measurable and that with the ordering on $\mathcal{M}$ that for all $A\in \mathcal{M}$, $A\subset \mathbb{R}^n$ (and therefore $A\prec \mathbb{R}^n$) and so $\mathcal{P}(\mathbb{R}^n)$, the power set of $\mathbb{R}^n$, contains a bound for $\mathcal M$.
Then, using Zorn's lemma, wouldn't we have that $\mathcal M$ contains the upper bound of the sequence $(T_i)_{i\in\mathbb{N}}$. Meaning that $$\bigcup_{j=1}^\infty S_j = \Big[\lim_{i\to\infty} T_i\Big]\in \mathcal{M}$$
Then we can do the same process using set containment as the ordering and $\emptyset$ as the bound to show that $\bigcap S_j$ is measurable.
Is my reasoning on this correct, or if not, where does my logic fail?
You really don't need Zorn's lemma for this. At best, you might want to make an appeal to the axiom of countable choice, but even that can be removed.
The main point is that the collection of all measurable sets is a $\sigma$-algebra, meaning it is closed under countable unions and intersections. Of course, this is what you have to show here. For this you need to make some sort of appeal to the way you defined measurable sets. Most commonly this is sets $A$ such that there is some Borel sets $A'$ and $B$ of measure zero such that $A\triangle A'\subseteq B$.
Then using countable choice, choose $S_j'$ and $B_j$ to be Borel sets for each $S_j$, then use the fact that Borel sets are closed under countable unions and intersections and the union of countably many measure zero Borel sets is a measure zero Borel set.
To your argument, you didn't prove that every chain has an upper bound in $\cal M$.
$\mathcal P(\Bbb R^n)$ is not an element of $\cal M$, it's a superset of $\cal M$, and it contains many non-measurable sets as well;
Zorn's lemma gives you a maximal element in $\cal M$, this is easy, $\Bbb R^n$. No need for Zorn's lemma. But this doesn't quite help you to show that the countable union of measurable sets is measurable.
Using Zorn's lemma usually means that the maximal element you obtain is an element with properties that solve your problem. Here the maximal element doesn't solve your problem at all, since any subset of $\Bbb R^n$ is a subset of $\Bbb R^n$, measurable or not.