Strange consequences of Axiom of Choice in Zermelo set theory

To sum up some of the discussion in the comments, I think you are greatly overestimating the role of Replacement in applications of the axiom of choice. The vast majority of transfinite induction arguments do not use Replacement in any essential way. One key to many applications is the following theorem of Hartogs (which is valid in Zermelo set theory without Choice):

Theorem: Let $X$ be a set. Then there exists a well-ordered set $W$ such that there is no injection $f:W\to X$.

Proof: Define $W$ to be the set of isomorphism classes of well-orderings of subsets of $X$, ordered by length. A well-ordering of a subset of $X$ is a subset of $X\times X$, and an isomorphism class is a set of such subsets, so $W\subset\mathcal{P}(\mathcal{P}(X\times X))$ and is a set by Separation. Note that the well-ordering of $W$ is longer than any element of $W$, since each element of $W$ is isomorphic to a proper initial segment of $W$ (namely, its own set of proper initial segments). If there existed an injection $f:W\to X$, the ordering on $W$ would give a well-ordering of the image $f(W)\subseteq X$ which has the same length as the well-ordering of $W$. But the well-ordering of $f(W)$ is an element of $W$. This is a contradiction.

This theorem can, for instance, be used to prove Zorn's lemma without Replacement. Given a poset $X$ which is a counterexample to Zorn's lemma, let $W$ be as in the theorem. Now just follow the usual transfinite recursion argument, using $W$ as the index set of the recursion rather than the ordinals. The argument constructs a strictly increasing function $W\to X$, which is in particular an injection. This is a contradiction.

By a similar argument, the theorem can be used to prove the well-ordering principle: using a choice function on nonempty subsets of $X$, just define a function $f:W\to X$ by transfinite recursion which is injective as long as possible (that is, $f(w)$ is different from $f(v)$ for all $v<w$ unless such values $f(v)$ are already all of $X$). Since $f$ cannot be an injection, this means that it must be surjective, and this gives a well-ordering of $X$.

You also mentioned in a comment:

I also find theorems that use the cardinality of the reals to be unintuitive, such as that there is a subset of the plane that intersects every straight line in exactly 2 points.

Results like these do not use Replacement either: they merely use a well-ordering of the reals. Given that a well-ordering of the reals exists, you can take one of minimal length, and then use that well-ordering as a substitute for the cardinality of the reals everywhere.