I'm trying to prove this well-known result. My proof is very simple that I suspect I made some subtle mistakes. Could you have a check on it?
Let $V$ be a linear space. A set $B \subseteq V$ is called a basis of $V$ if each $x\in V$ is a finite linear combination of elements in $B$ and if each finite subset of $B$ is linearly independent. By axiom of choice, such $B$ always exists. Then $\dim V := \operatorname{card} B$ is independent of the choice of $B$.
My attempt: Let $C$ be another basis of $V$. We now define a map $T:B \to C$.
For each $b \in B$, $\mathbb Rb$ is a vector subspace of $V$, then there is $c_b\in C$ such that $\mathbb Rc_b=\mathbb Rb$. Notice that $c_b$ is linearly independent of each element in $C\setminus\{c_b\}$. Hence such $c_b$ is unique. We define $T(b):=c_b$.
Assume that $T(b_1)=T(b_2)$. Then $\mathbb Rb_1 =\mathbb Rb_2$. Notice that $b_1$ is linearly independent of each element in $B\setminus\{b_1\}$, so $b_1=b_2$ and thus $T$ is injective. It follows that $\operatorname{card} B \le \operatorname{card} C$.
By symmetry, we have $\operatorname{card} C \le \operatorname{card} B$. This completes the proof.
Update: As mentioned in the comment, above proof is not correct. I fix it as follows.
WLOG, we assume $B$ and $C$ are infinite. Let $F_B, F_C$ denote the collections of all finite subsets of $B$ and $C$ respectively. By axiom of choice, there is a map $T:B \to F_C$ such that $b$ is a linear combination of vectors in $T(b)$. We have $$V=\operatorname{span} B \subseteq \operatorname{span} \left ( \bigcup_{b\in B} T(b) \right )\subseteq V.$$
Because $C$ is linearly independent, it is the only subset of itself that spans $V$. So $$\bigcup_{b\in B} T(b) = C.$$
We have $$|C| = \left | \bigcup_{b\in B} T(b) \right | \le |B| \cdot \aleph_0 = |B|.$$ By symmetry, $|B| \le |C|$. This completes the proof.
