Bit of a spoiler: Your approach seems on the way to what I've seen done, but instead of trying to intersect your set, you might want to map a non measurable one into it using a measurable map and remember how preimages of borel sets behave.
Spoiler: your map could be one from the unit interval onto that very famous set by that very famous guy born in 1845 who suffered from depression and the dislike of many of his contemporaries... ;-)
First of all, for many people, using the "Lebesgue measure" does not imply that you are dealing with all "Lebesgue measurable" sets. The measure in $\mathbb{R}$ that satisfies $\mu((a,b]) = b - a$ is usually called Lebesgue measure only when it is completed. But some may call the measure restricted to the Borel sets with the same name.
The statement
In fact, practically any function that can be described is measurable.
is not really meaningful. It depends a lot on one's ability to describe/construct a function.
About the "composition of functions", it is important to notice that if you are talking about measurable spaces, then the functions do not simply take one set to another... they take one space to another space. Let's write $\mathcal{B}$ for the Borel sets and $\mathcal{L}$ for the Lebesgue measurable sets. The problem is that very often probabilists say that a function $f: \mathbb{R} \to \mathbb{R}$ is measurable when $f: (\mathbb{R}, \mathcal{L}) \to (\mathbb{R}, \mathcal{B})$ is measurable. But, in general, when dealing with $f: (\Omega_1, \sigma_1) \to (\Omega_2, \sigma_2)$ and
$g: (\Omega_2, \sigma_2) \to (\Omega_3, \sigma_3)$, it IS TRUE that
$g \circ f: (\Omega_1, \sigma_1) \to (\Omega_3, \sigma_3)$ IS measurable.
Probably, they do that because they are used to defining $f: (\Omega, \sigma) \to \mathbb{R}$ as measurable when $f^{-1}([a,b]) \in \sigma$ for all $a,b \in \mathbb{R}$.
About your last question, I guess it is clear that in many cases it is easier to deal with a complete space. Nevertheless, notice that the concept of "completion" does depend on a fixed measure. The Borel sets is a family that depends only on the topology, while you do need a measure in order to complete it. So, if you are talking about all measures that one may define on a certain topological space, or on a certain measurable space, you will not talk about the completion of the space. One example is the Riesz representation theorem.
Best Answer
The basic reason is that a Lebesgue measurable function is one with the property that preimages of Borel sets are Lebesgue measurable. The main reason why we define things that way is because not all continuous functions are Lebesgue-Lebesgue measurable, so if we defined measurability through Lebesgue-Lebesgue measurability then the Lebesgue integral would not be an extension of the Riemann integral.
Anyway, the structural reason why this has anything to do with composition is because:
$$(f \circ g)^{-1}(A)=g^{-1}(f^{-1}(A)).$$
Thus if $A$ is Borel and $f,g$ are Lebesgue-Borel measurable, then $f^{-1}(A)$ might be Lebesgue measurable but not Borel measurable, and so $g^{-1}(f^{-1}(A))$ could fail to even be Lebesgue measurable. To do this construction, it suffices to take $g$ to be a Lebesgue-Borel measurable function which is not Lebesgue-Lebesgue measurable, find a set $M$ which is Lebesgue measurable such that $g^{-1}(M)$ is not Lebesgue measurable, and then set $f=1_M,A=\{ 1 \}$.
As you may be aware, making up Lebesgue nonmeasurable sets is never really "constructive", but there is still a way to build such a $g$ and show it has the necessary properties, provided we can conjure up a suitable Lebesgue nonmeasurable set in the course of the proof. To do that, we look at
$$h : [0,1] \to [0,2], h(x)=c(x)+x$$
where $c$ is the standard Cantor function. It is easy to see that $h$ is a strictly increasing continuous function. It's perhaps less easy to see that $\mu(h(C))=1$ where $C$ is the standard Cantor set; you might call this lemma 1. With that in hand, you can select a Lebesgue nonmeasurable set $N \subset h(C)$ (you might call this ability to construct a nonmeasurable subset of a positive measure set lemma 2).
Now $h^{-1}(N)$ is a Lebesgue measurable set (because it is a subset of $C$). But its preimage under $h^{-1}$ (i.e. $N$) isn't Lebesgue measurable. It turns out this is only possible because it is not Borel, because $h(B)$ is Borel for every Borel $B$ since $h$ is strictly increasing.
In the end we can take $g=h^{-1},M=h^{-1}(N)$ in the notation of the third paragraph.
I wrote this answer essentially by following along with https://www.math3ma.com/blog/lebesgue-but-not-borel but the answer is complete even if this link dies.