Your argument doesn't work, since - as mentioned in the comments - given any infinite set $A$ there is a map $f:A\rightarrow A$ which is surjective but not injective (or injective but not surjective, if you prefer).
However, the statement is true, and the proof is simpler:
Suppose $B, B'$ are infinite bases of $\mathbb{V}$ with $\vert B\vert<\vert B'\vert$.
For each $b\in B$, there is a finite $F_b\subseteq B'$ with $b\in\langle F_b\rangle$.
Since the union of $\vert B\vert$-many finite sets has cardinality $\vert B\vert$, the set $$\hat{B}=\bigcup_{b\in B}F_b$$ has cardinality $\vert B\vert$.
Since $\vert B\vert<\vert B'\vert$, there must therefore be some $a\in B'\setminus\hat{B}$.
But since $B$ spans $\mathbb{V}$, $a$ can be written as a linear combination of elements of $B$, and hence can be written as a linear combination of elements of $\hat{B}$. This contradicts the linear independence of $B'$.
As a quick pedagogical aside, here's an attempted proof which doesn't quite work:
Since $B$ is a basis for $\mathbb{V}$, we have that every element of $B'$ can be represented as a finite linear combination of elements of the field of scalars $F$. Using the fact that the set of finite subsets of an infinite set has the same cardinality as the original set, the cardinality of the set of linear combinations of elements of $B$ is $\vert B\vert\times\vert F\vert$, and so we get $$\vert B'\vert\le\vert B\vert\times\vert F\vert.$$
In case $F$ is "small" - that is, in case $\vert F\vert\le\vert B\vert$ - then we get $\vert B'\vert\le\vert B\vert$ (since multiplication of two infinite cardinalities just results in the larger of the two). But in case $F$ is large, we don't get anything. E.g. this argument doesn't rule out the possibility of a vector space over a field of cardinality $\aleph_{\omega^3+\omega\cdot 842+17}$ with one basis of cardinality $\aleph_{\omega^2+9}$ and another basis of cardinality $\aleph_0$.
Note that we've used the axiom of choice crucially in the above when we calculated the cardinality of $\hat{B}$ (which we needed in order to conclude the existence of $a$). Without the axiom of choice, the argument above breaks down. This is the only essential use of choice - note that the second bulletpoint does not require choice, since there is a unique minimal choice of $F_b$.
The full axiom of choice is not needed, and the general study of how much choice is needed to prove various mathematical results - around vector spaces and other topics - is quite rich; I just want to point out the reliance on something beyond set theory without choice, here.
