(This is more of a comment than an answer, but I think the following caveat is indispensable whenever Cauchy reals are mentioned in constructive mathematics without Countable Choice.)
I'm puzzled by the fact that no answer to this question so far seems to mention the fact that (in the absence of Countable Choice) Cauchy reals can fail to be Cauchy-complete, so let me point to the paper “On the Cauchy Completeness of the Constructive Cauchy Reals” by Robert Lubarsky which explains this perplexing state of affairs well. (In contrast, the Dedekind reals are Cauchy-complete, and they are also Dedekind-complete.)
The situation about Cauchy reals is made very confused that, depending on authors, a “Cauchy sequence” of rationals may be a sequence $(r_n)$ such that $\forall N. \exists n. \forall p,q\geq n. (|r_p-r_q| < 2^{-N})$ or it may be one such that there exists $\nu\colon\mathbb{N}\to\mathbb{N}$ such that $\forall N. \forall p,q\geq \nu(N). (|r_p-r_q| < 2^{-N})$ (i.e., requiring a modulus of convergence). The two notions fail to coincide, so we actually have at least two different notions of Cauchy reals¹: Cauchy-reals-without-modulus and Cauchy-reals-with-modulus. When considering Cauchy sequences of Cauchy sequences, for the outer sequence we can demand a modulus or not, and for the inner sequences we can demand existence of a modulus for each one separately or the existence of a sequence of modulus sequences, and pretty much everything that can go wrong does; but in any case, a Cauchy real (with or without modulus) should be taken to be an equivalence class of Cauchy sequences of rationals (with or without modulus), and the main problem is that there is no way to choose a representative from each class in the sequence. This is all proved in Lubarsky's paper.
I don't know what happens if we define “Cauchy reals” as the smallest subset² of the Dedekind reals which contains the rationals and is closed under taking Cauchy sequences (with or without modulus). This would give two more notions of “Cauchy” reals to add to the mess.
One piece of good news, however, is that if the difference between two Cauchy-reals-with-modulus converges-to-zero-without-modulus (viꝫ. $\forall N. \exists n. \forall k\geq n. (|r_k-s_k| < 2^{-N})$ where $(r_n)$ and $(s_n)$ are the two sequences), then it even converges-to-zero-with-modulus (viꝫ. there is $\mu\colon\mathbb{N}\to\mathbb{N}$ such that $\forall N. \forall k\geq \mu(N). (|r_k-s_k| < 2^{-N})$. This is very easy, but rarely well explained. So we don't have to distinguish two different equivalence relations on Cauchy-reals-with-modulus.
I think this definition makes sense, but at this point I'm so terrified of lions lurking in the shadows that I wouldn't bet my hand on it.