An answer to your question can be given in intuitionistic logic. If you are not a fan of intuitionistic logic and don't know much about it; it's enough to know that the law of excluded middle ($\mathsf{LEM}$) can not be shown in said logic.

What you can however show in intuitionistic logic is that $\mathsf{LEM}$ and and Contraposition ($\mathsf{CP}$) are equivalent and in this precise sense, $\mathsf{CP}$ is "just as bad" a $\mathsf{LEM}$.

Let me expand on my answer, more specifically addressing the bullet-points.

Yes there is a proof of the equivalence between $\mathsf{LEM}$ and $\mathsf{CP}$ which, upon inspection, does not use double negation elimination ($\mathsf{DNE}$). Actually, from the start, we can slightly change our classical proof system and completely remove the possibility to use $\mathsf{DNE}$, which guarantees that we don't use it by accident. There is even a name for this variation of classical logic and you might have guessed it by now: it's intuitionistic logic.

Yes. There are good reasons for people to avoid proofs which use $\mathsf{LEM}$ and since it is equivalent to $\mathsf{CP}$, this also means we are trying to avoide $\mathsf{CP}$.

Yes there are instances where $(\neg B \rightarrow \neg A) \rightarrow (A \rightarrow B)$ can be proven and therefore be used, just as there are instances of a statement that has the form $X \lor \neg X$ which can be proven and therefore be used. For example, if you have the finite set $B = \{0, 1\}$ you can (without using $\mathsf{LEM}$ or the likes) show that $\forall x, y \in B : ~ x = y \lor \neg (x = y)$.

Here are proofs for $\mathsf{CP} \Rightarrow \mathsf{DNE} \Rightarrow \mathsf{LEM}$ and lastly $\mathsf{LEM} \Rightarrow \mathsf{CP}$ in intuitionistic logic. Rember that in intuitionistic logic (just like in classical logic) there are the always false proposition $\bot$ and the always true proposition $\top$ and negation is defined by $\neg P := P \rightarrow \bot$.

Assume $\mathsf{CP}$. This means for any propositions $A, B$ we have $(\neg B \rightarrow \neg A) \rightarrow (A \rightarrow B)$.
So in particular for $A = \neg \neg X$ and $B = X$ we get
$$
(\neg X \rightarrow \neg \neg \neg X) \rightarrow (\neg \neg X \rightarrow X)
$$
So $\neg \neg X \rightarrow X$ follows from $\neg X \rightarrow \neg \neg \neg X$ where this latter statement is true since by definition it is the same as $(\neg X) \rightarrow \neg (\neg X) \rightarrow \bot$ which is more obviously true.

For the second implication we now assume $\mathsf{DNE}$. This means for every $A$ we have $\neg \neg A \rightarrow A$, which means in order to show $X \lor \neg X$ it suffices to show $\neg \neg (X \lor \neg X)$.
So assume $H_1 : \neg (X \lor \neg X)$ and try to get a contradiction to it. With $H_1$ we can show $H_2 : \neg X$ (since assume we had $X$, then we also have $X \lor \neg X$; in contradiction to $H_1$). Now that we have $H_2$ this too implies $X \lor \neg X$ and we therefore get the desired contradiction to $H_1$.

Assume $\mathsf{LEM}$, meaning for every $X$ we have $X \lor \neg X$. We want to show $(\neg B \rightarrow \neg A) \rightarrow (A \rightarrow B)$, so we assume $H_1 : (\neg B \rightarrow \neg A)$ and $H_2 : A$ and try to show $B$. Because of $\mathsf{LEM}$ we either have $B$ or $\neg B$. In the first case it is trivial to show $B$. In the second case we have $\neg B$ which, by using $H_1$ gets us $\neg A$ which by $H_2$ gets us $\bot$. So by explosion we can again conclude $B$.

By request fom the comments, here is also the implication $\mathsf{LEM} \Rightarrow \mathsf{DNE}$.

We are trying to show $\neg \neg X \to X$, so assume $H : \neg \neg X$ and try to get $X$. By $\mathsf{LEM}$ we know that $X \lor \neg X$ so we get two cases:

- If $X$ then we are done.
- If $\neg X$ then we get $\bot$ because of $H$ and therefore $X$ by explosion.