linear algebra - explicit big linearly independent sets

Here is a linearly independent subset of $mathbb{R}$ with size $2^{aleph_0}$.

Let $q_0, q_1, ldots$ be an enumeration of $mathbb{Q}$. For every real number $r$, let
$$T_r = sum_{q_n < r} frac{1}{n!}$$
The proof that these numbers are linearly independent is similar to the usual proof that $e$ is irrational. (It's a cute problem; there's spoiler below.)

I think a similar trick might work for algebraic independence, but I don't recall having seen such a construction. Actually, John von Neumann showed that the numbers
$$A_r = sum_{n=0}^infty frac{2^{2^{[nr]}}}{2^{2^{n^2}}}$$
are algebraically independent for $r > 0$. [Ein System algebraisch unabhängiger zahlen, Math. Ann. 99 (1928), no. 1, 134–141.] A more general result due to Jan Mycielski seems to go through in ZF + DC perhaps just ZF in some cases. [Independent sets in topological algebras, Fund. Math. 55 (1964), 139–147.]

As for subspaces and subfields isomorphic to $mathbb{R}$, the answer is no. (Since I'm not allowed to post any logic here, I'll refer you to this answer and let you figure it out.)

Well, I'll bend the rules a little... Consider a $mathbb{Q}$-linear isomorphism $h:mathbb{R}to H$, where $H$ is a $mathbb{Q}$-linear subspace of $mathbb{R}$ (i.e. $h$ is an additive group isomorphism onto the divisible subgroup $H$ of $mathbb{R}$). If $h$ Baire measurable then it must be continuous by an ancient theorem of Banach and Pettis. It follows that $h(x) = xh(1)$ for all $x in mathbb{R}$ and therefore $H = mathbb{R}$. Shelah has produced a model of ZF + DC where all sets of reals have the Baire property, so any such $h$ in this model must be Baire measurable. A similar argument works if Baire measurable is replaced by Lebesgue measurable, but Solovay's model of ZF + DC where all sets of reals are Lebesgue measurable uses the existence of an inaccessible cardinal, and this hypothesis was shown necessary by Shelah.

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Spoiler

Suppose for the sake of contradiction that $r_1 > r_2 > cdots > r_k$ and $a_1,a_2,ldots,a_k in mathbb{Z}$ are such that $a_1T_{r_1} + a_2T_{r_2} + cdots + a_kT_{r_k} = 0$. Choose a very large $n$ such that $r_1 > q_n > r_2$. If $n$ is large enough that
$$(|a_1| + |a_2| + cdots + |a_k|) sum_{m=n+1}^infty frac{n!}{m!} < 1$$
then the tail terms of $n!(a_1T_{r_1}+cdots+a_kT_{r_k}) = 0$ must cancel out, and we're left with
$$a_1 = -sum_{m=0}^{n-1} sum_{q_m < r_i} a_i frac{n!}{m!} equiv 0 pmod{n}$$
If moreover $n > |a_1|$, this means that $a_1 = 0$. Repeat to conclude that $a_1 = a_2 = cdots a_k = 0$.