Let $V$ be a vector space over $\mathbb{R}$ with basis $\mathcal{F}, $ such that $|\mathcal{F}| = 2^{\aleph_0}$. Prove that $|V| = 2^{\aleph_0}.$
Clearly, $|\mathbb{R}| \leq |V|$ as the map $k\mapsto kb,$ where $ b\in\mathcal{F},$ is injective. Since $V$ is a vector space over $\mathcal{F},$ any element of $V$ can be written as a unique finite linear combination of elements of $\mathcal{F}$ (with zero coefficients ignored, unless $v=0$, in which case $0=0 b_1$ for any $b_1\in \mathcal{F}$). Thus we can map each element $v\in V$ to the finite subset $S_v$ of the set $ \mathcal{F}\times \mathbb{R}$ such that if $\alpha \neq 0$ is the corresponding coefficient of $b\in\mathcal{F}$ in the unique linear combination corresponding to $v,$ then $(b,\alpha)\in S_v$ (so if $v = 0,$ map $v$ to the subset $\emptyset$). This map is indeed injective, as if $S_v = S_w $ for some $v, w\in V,$ then the linear combination corresponding to $S_v,$ which is $v,$ equals the linear combination corresponding to $S_w,$ which is $w.$ Observe that $|\mathcal{F}\times \mathbb{R}| = |\mathcal{F}||\mathbb{R}| = 2^{\aleph_0} 2^{\aleph_0} = 2^{\aleph_0+\aleph_0} = 2^{\aleph_0}.$ For a set $X,$ let $S(X)$ denote the set of finite subsets of $X$. Then if $X$ is infinite, one can show $|S(X)| = |X|.$
Thus $S(\mathcal{F}\times \mathbb{R}) = |\mathcal{F}\times \mathbb{R}| = 2^{\aleph_0}.$ So $|V|\leq |\mathcal{F}\times \mathbb{R}| = 2^{\aleph_0}$ by the described injection $v\mapsto S_v$. Thus, $|V|\leq |\mathbb{R}|.$ By the Cantor-Schroedner-Bernstein theorem, $|V| = |\mathbb{R}|.$
Your proof is correct. Here are some suggestions.
You do not need the map $k\mapsto kb,$ where $ b\in\mathcal{F}$, which is injective. This part is covered by your argument that there is a bijection between $V$ and the set $S(\mathcal{F}\times \mathbb{R})$ of finite subsets of $\mathcal{F}\times \mathbb{R}$. Note that $0 \in \mathbb R$ corresponds to $\emptyset \in S(\mathcal{F}\times \mathbb{R})$ because one defines the sum over the empty index set to be $0$. This convention is also reflected in the statement that the trivial vector space $\{0\}$ has a basis (this is $\emptyset$).
Now you know that $V$ and $S(\mathcal{F}\times \mathbb{R})$ have the same cardinality and you have determined $\lvert S(\mathcal{F}\times \mathbb{R}) \rvert = 2^{\aleph_0}$.