Are two finite-dimensional subspaces not disjoint iff the union of any of their bases is linearly dependent? First I need to show the following Lemma:
Lemma
Let $\mathcal{M},\mathcal{N}$ be finite-dimensional subspaces of $V$ with any bases $\mathcal{M}_{0}$ and $\mathcal{N}_{0}$. Then, $\mathcal{M}\cap\mathcal{N}\neq \{0\}$ if and only if $\mathcal{M}_{0}\cup \mathcal{N}_{0}$ is linearly dependent.
Proof. Consider any bases $\mathcal{M}_{0},\mathcal{N}_{0}$ such that their union is linearly dependent. Then, $\sum\alpha_{i} m_{i} + \sum\beta_{i}n_{i}=0$ implies that, at least, $\alpha_{j},\beta_{i}\neq 0$ for some $i,j$. Thus, $\sum \alpha_{i} m_{i} = -\sum \beta_{i} n_{i} \neq 0$, which implies that a nonzero linear combination in $\mathcal{M}$ can be expressed as a linear combination in $\mathcal{N}$ and so $\mathcal{M}\cap\mathcal{N} \neq \{0\}$.
For the converse, assume that $\mathcal{M}\cap \mathcal{N}\neq \{0\}$. Hence, there is a nonzero vector $c\in \mathcal{M}\cap\mathcal{N}$ and so $c = \sum \alpha_{i}m_{i} = \sum\beta_{i}n_{i}$, where, at least, $\alpha_{i},\beta_{j} \neq 0$. Hence, $\sum\alpha_{i}m_{i} - \sum \beta_{i}n_{i} = 0$ and so $\mathcal{M}_{0}\cup\mathcal{N}_{0} $ is linearly dependent.
Q.E.D
Now, consider two nontrivial subspaces $\mathcal{M}$ and $\mathcal{N}$ of $V$. In the case that $\mathcal{M}\cap \mathcal{N}\neq \{0\}$, there is a linearly independent set $C_{0}$ that spans $\mathcal{M}\cap\mathcal{N}$ and so one can extend it to bases $\mathcal{M}_{0}$ and $\mathcal{N}_{0}$. Hence, $M_{0}\cap\mathcal{N}_{0}=C_{0}$ and $\mathcal{M}_{0}\cup\mathcal{N}_{0}$ is linearly dependent. However, I have seen in some proofs of the equality $\text{dim} \mathcal{M} + \text{dim} \mathcal{N} = \text{dim}(\mathcal{M}+\mathcal{N}) + \text{dim}(\mathcal{M}\cap \mathcal{N})$ that $\mathcal{M}_{0}\cup \mathcal{N}_{0}$ is a basis of $\mathcal{M}+\mathcal{N}$, namely, $\mathcal{M}_{0}\cup \mathcal{N}_{0}$ is linearly independent. There must be something wrong with my proof, hoever I'm pooped and need some help.
Thanks
 A: 
Lemma
Let $\mathcal{M},\mathcal{N}$ be finite-dimensional subspaces of $V$ with any bases $\mathcal{M}_{0}$ and $\mathcal{N}_{0}$. Then, $\mathcal{M}\cap\mathcal{N}\neq \{0\}$ if and only if $\mathcal{M}_{0}\cup \mathcal{N}_{0}$ is linearly dependent.

The lemma above is correct only if we interpret $\mathcal{M}_0\cup\mathcal N_0$ as the union of $\mathcal M_0$ and $\mathcal N_0$ as multisets, which is somewhat unconventional since "$\cup$" is used to denote the union of sets by default usually. For example, if $\mathcal{M}_{0}=\mathcal{N}_{0}=\{v\}$, then $\mathcal{M}_{0}\cup \mathcal{N}_{0}$ means the multiset $[v,v]$, where $v$ is included two times.
Suppose we do interpret $\mathcal{M}_0\cup\mathcal N_0$ as the union of $\mathcal M_0$ and $\mathcal N_0$ as multisets. While your proof of the lemma cannot be marked as wrong, it lacks details as Arturo Magidin pointed out in their comment.

On the other hand,  "in some proofs of the equality $\text{dim} \mathcal{M} + \text{dim} \mathcal{N} = \text{dim}(\mathcal{M}+\mathcal{N}) + \text{dim}(\mathcal{M}\cap \mathcal{N})$", "$\mathcal{M}_{0}\cup \mathcal{N}_{0}$ is a basis of $\mathcal{M}+\mathcal{N}$, namely, $\mathcal{M}_{0}\cup \mathcal{N}_{0}$ is linearly independent." I would assume in those proofs, $\mathcal{M}_{0}\cup \mathcal{N}_{0}$ is used to denote the union of $\mathcal{M}_{0}$ and $\mathcal{N}_{0}$ as sets. For example, if $\mathcal{M}_{0}=\mathcal{N}_{0}=\{v\}$, $\mathcal{M}_{0}\cup \mathcal{N}_{0}$ will be the set $\{v\}$.
