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In the comments to this answer there are a few statements that interest me which I would like some clarification on. Apologies if the material referred to has been asked about here before. Asaf Karagila makes the comment:

As I keep pointing out, rejecting choice and taking up "all sets are measurable" implies that you can partition the reals into more non-empty parts than points. If that is not paradoxical, I don't know what is...

and then explains this in a later comment as follows:

It was shown by Sierpinski (if my memory serves me right at this time of night) that if there is a bijection between $\mathbb{R}$ and $[\mathbb{R}]^\omega$ (the set of all countably infinite subsets of $\mathbb{R}$), then there is a non-measurable set. Note that there is always an injection from $\mathbb{R}$ into that set, and there is always a surjection from $\mathbb{R}$ onto that set. So if all sets are measurable, we can partition $\mathbb{R}$ into $|[\mathbb{R}]^\omega|$ different, non-empty parts, which is strictly more than $|\mathbb{R}|$. Simply fix such a surjection, and look at its fibers.

I do not know the source of the theorem of Sierpinski referred to here, and I would love it if someone could point me towards it. I think I understand the rest of the argument; I assume since an injection (e.g. map each $x\in\mathbb{R}$ to its singleton?) and surjection can exist without a bijection that Cantor-Schroder-Bernstein is being rejected here; I think I get why we can partition $\mathbb{R}$ into $|[\mathbb{R}]^\omega|$ parts if there is a surjection from $\mathbb{R}$ onto $[\mathbb{R}]^\omega$ (for each element in $[\mathbb{R}]^\omega$ we find the corresponding fiber of the surjection), and I see why $|[\mathbb{R}]^\omega|>|\mathbb{R}|$ (since there is an injection from $\mathbb{R}$ into $[\mathbb{R}]^\omega$ without a bijection; please correct my reasoning if it wrong). However, I don't quite see why the surjection from $\mathbb{R}$ to $[\mathbb{R}]^\omega$ must exist.

Thus I have the following questions: Could anyone point me to the theorem of Sierpinski (perhaps) referred to here, and could anyone point out why the surjection from $\mathbb{R}$ to $[\mathbb{R}]^\omega$ must exist?

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    The Cantor-Schröder-Bernstein theorem says that, if there is an ***INJECTION*** from $A$ to $B$ and an ***INJECTION*** from $B$ to $A,$ then there is a bijection between $A$ and $B.$ The theorem does not apply when you have an injection and a ***SURJECTION*** from $A$ to $B.$ – 2017-02-06
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    For instance, if $A=\mathbb R$ and $B=\mathbb R\cup\omega_1,$ then (without the axiom of choice) you have an injection from $A$ to $B,$ and a surjection from $A$ to $B,$ and a surjection from $B$ to $A,$ but you need the axiom of choice to get a bijection from $A$ to $B.$ The Cantor-Schröder-Bernstein theorem is no help here. – 2017-02-06
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    @bof True that. My mistake. – 2017-02-06

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Note that $\Bbb R$ and $\Bbb R^\omega$ are equipotent. Now map every sequence to its image (add the rationals if that image is finite).

You can find a discussion with citation and proof in this MathOverflow thread.

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    Thank you for the reply, but there's something I'm not getting here. I thought two sets being equipotent was equivalent to saying there was a bijection between them, but I thought when we were assuming here that all sets were measurable that implied that there was no bijection (by the quoted theorem) between $\mathbb{R}$ and $[\mathbb{R}]^\omega$ (and hence the size of the latter set is strictly greater than that of the former). Is there another meaning of equipotent that I've missed? – 2017-02-06
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    @Anon $[\mathbb R]^\omega\ne\mathbb R^\omega$ – 2017-02-06
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    @bof My mistake again, I shouldn't have commented so quickly. Thank you. That answers my second question. Now the only other thing is to find the source of that theorem. – 2017-02-06
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    @Anon: I've added a link to an old MO question by yours truly. – 2017-02-06