I am supposed to prove or disprove that two linearly independent subsets $V_1$ and $V_2$ of a finite-dimensional vector space $V$ having same span (that is $\text{span}(V_1) = \text{span}(V_2)$) have equal cardinality. Here is my attempt at solving this problem, please grade the solution.
Solution: We begin with a claim :
Claim: If $V_i$ is a subset of a vector space $V$, then $\text{span}(V_i)$ is the smallest subspace of $V$ containing all elements of $V_i$, that is any such subspace $U$ containing all elements of $V_i$ must satisfy $U \supseteq \text{span}(V)$.
Proof: Just observe that $\text{span}(V_i)$ indeed contains all elements of $V_i$. But any subspace $U$ containing $V_i$ must contain any linear combination of elements of $V_i$ which is $\text{span}(V_i)$ so it suffices to show that $\text{span}(V)$ is a subspace, which is trivial as it is closed under addition and scalar multiplication.
Now back to the proof, observe that since $\text{span}(V_1) = \text{span}(V_2) = U$, and $V_1, V_2$ are linearly independent, it follows that $V_1, V_2$ are bases of $U$. But given any subspace $U$, using the lemma that length of spanning list is $\geq$ length of linearly independent subset, we obtain that two bases of a given subspace $U$ have the same cardinality. Thus, $V_1, V_2$ have the same cardinality, as desired.