Is the reverse of a closed under operation maintainable?
No, if $L_1\cup L_2$ is decidable $L_1$ and $L_2$ are not necessarily decidable.

Here's an example.

$$\begin{align} 
L_1 &amp;= \text{any language that is not decidable.}\\
L_2 &amp;= \Sigma^*\setminus L_1.
\end{align}$$
$L_1\cup L_2$ is $\Sigma^*$, which is decidable. However, $L_1$ is not decidable. In fact, $L_2$ is not decidable, either.



The proof given in the preformatted block of the question is a proof for the original proposition, "if $L_1$ and $L_2$ are decidable,  then $L_1\cup L_2$ is decidable". It is not a proof for its converse proposition.

To prove its converse proposition, all you can assume is a Turing machines that decides $L_1\cup L_2$. There is no way that you can construct a Turing machine that can decide $L_1$ from that machine.



Intuitively, one condition cannot imply two other conditions. That is, "$L_1 \cup L_2$ is decidable" cannot imply both "$L_1$ is decidable" and "$L_2$ is decidable"

Here are a few related exercises.

Exercise 1. Does $L_1 \cup L_2$ is regular imply $L_1$ is regular and $L_2$ is regular?

Exercise 2. Does $L_1 \cup L_2$ is context-free imply $L_1$ is context-free and $L_2$ is context-free?

Exercise 3. Does $L_1 \cup L_2$ is recursively enumerable imply $L_1$ is recursively enumerable and $L_2$ is recursively enumerable?

Exercise 4. Raise a question similar to the above question.

Exercise 5. If you know that $L_1\cup L_2$ and $L_1$ are both decidable, does that imply $L_2$ is decidable?





The case with intersection is the same. Please check question if $L_1$ and $L_2$ are languages over the same alphabet and $L_1\cup L_2$ is context free, at least one of them must be context free, its answer and its exercises.