Here's the question:
Let $V$ be a vector space over $\mathbb R$ of dimension $8$. Let $V_1, V_2, V_3$ be subspaces of $V$ such that $\dim V_i =k$ for $i=1,2,3$ and $V_1\cap V_2 \cap V_3 = \{0 \}$. What is the largest value $k$ can have?
My attempt
It is easy to see that $\dim (V_1 \cap V_2 \cap V_3) \le \min_{1\le i,j \le 3} \{ \dim (V_i \cap V_j) \}$. If we have that $V_i \cap V_j =\{ 0 \}$ for some $i\ne j$ then we would be done and in such a case, $\dim (V_i + V_j)=2k$ and so the largest $k$ that we can have is $k=4$.
Now if $k\ge5$ then $\dim (V_1 \cap V_2)=\dim (V_1 + V_2) - \dim V_1 - \dim V_2 = \dim (V_1 + V_2) -2k\le 8-2k$ which is not possible.
Is my argument correct? Alternative approaches are welcome!
Your argument is corret insofar it proves that $k$ cannot be greater or equal than $5$.
Your argument that $k$ can be $4$, however, is lacking. Your argument really only says "if $k=4$, then the dimensions of $V_i$ and $V_j$ match up and that's fine", but you didn't really demonstrate anything. Just because the dimension of $V_i+V_j$ is $2k$, and that is $8$, you did not yet show that the third space, $V_l$ where $l\neq i$ and $l\neq j$, will not cause problems.
The best way to prove that $k=4$ is possible is to actually construct $V_1,V_2,V_3$ such that they all have dimension $4$ and their intersection is $\{0\}$.