Regarding the Post: "Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?"

You don't need an indirect argument (that is, saying "for otherwise ..." is a detour). Just use the property directly:

$\mathbb R$ is the disjoint union of $I$ and $\mathbb Q$, and I assume you know $|\mathbb Q|=\aleph_0$, so this tells you $|I|+\aleph_0 = \mathfrak c$. Therefore, setting $k=|I|$ in the property you link to, we get $|I|=\mathfrak c$. Done!

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Alternatively, you can simply describe a concrete bijection $f:\mathbb R\to I$:

$$ f(x) = \begin{cases} x+\sqrt2 & \text{if there is an }n\ge 0\text{ such that }x-n\sqrt2\in\mathbb Q \\ x & \text{otherwise} \end{cases} $$

You are correct, and the result does not require the axiom of choice as was proved in that other post; however your proof uses the axiom of choice, the way that it is stated.

The proof cited in the post works for any cardinal $\mathfrak{m}$ and without the axiom of choice. However here you are in a more specific case ($\mathfrak{m} = \aleph_0$) and so you can do a more specific proof.

For instance, you can use continued fractions to produce an explicit injection $\mathbb N^{\mathbb N}\to I$ which in particular implies the desired result.

Or you can also argue that $I$ is a $G_\delta$-set, therefore it satisfies the continuum hypothesis : it has cardinality either $\aleph_0$ or $2^{\aleph_0}$, and your argument (this time without choice) shows that it can't be $\aleph_0$