Determine whether $L$ defined by $Lu = u' + |u|$ is a linear operator.
I claim yes, but only in certain cases.
Proof
Suppose $L$ is a linear operator then by definition
$$L(av + bw) = (av + bw)' + \left|av + bw\right| = a\left(v' + \left|v\right|\right) + b\left(w' + \left|w\right|\right) = aLv + bLw$$
which implies
$$\left|av + bw\right| = a\left|v\right| + b\left|w\right| \tag{$\star$}$$
Applying the triangle inequality to the left side of $\,(\star)$
$$ \left|av + bw\right| \leq \left|a\right|\left|v\right| + \left|b\right|\left|w\right|$$
So this contradicts the supposition that $L$ is linear in general.
Looking back I believe I should reconstruct the proof in two parts: (i) proving $L$ is linear with the restrictions that $\,a,b\geq0\, $and that $v(x), w(x)$ are the same sign with respect to each other and that that sign never changes (but can be zero). (ii) for the second part I will redo the above proof supposing $L$ is linear for all other cases not subject to the restrictions of part (i) and arrive at a contradiction.
Please comment on my approach and validity, because to me it seems like overkill. Is there a non case by case method?
First of all, your proof as written is not correct: you haven't reached any contradiction. Apparently you seem to think the inequality you found is somehow a contradiction but it is not in any obvious way.
Second of all, for $L$ to be linear means that $L(av+bw)=aL(v)+bL(w)$ in all cases. If this equation is true in some cases but not others, then $L$ is simply not linear, rather than "linear in some cases". Moreover, that means that to answer the question, you just need to find one case where it fails: that is all it takes to conclude that $L$ is not linear.
So your goal is just to give one counterexample. That is, you just need to describe one way to choose $a,b,v,$ and $w$ such that $L(av+bw)\neq aL(v)+bL(w)$. The easiest and best way to do this is be completely explicit: actually say what specific numbers $a$ and $b$ are and what functions $v$ and $w$ are, so that you can compute $L(av+bw)$ and $aL(v)+bL(w)$ and see they are not the same. Using the work you've done so far, it would be enough to just check that $|av+bw|\neq a|v|+b|w|$ in your example.
The casework you propose doing is not helpful. It is not relevant whether $L(av+bw)=aL(v)+bL(w)$ in some cases (as in your case (i)); the only thing you care about proving in this problem is that there exists a case where it isn't true. (Of course, the fact that $L(av+bw)=aL(v)+bL(w)$ in some cases may be useful in other contexts, if you are interested in more than just this single problem.)