Let say K and L are two different subspace real vector space V. If given...

Explanation:
Let the four vectors ${\displaystyle k_1,k_2,k_3}$ and ${\displaystyle k_4}$ form a basis of the vector space K. Since K is a subspace of V, these four vectors form a linearly independent set in V. Since L is a subspace of V different from K, there must be at least one element, say ${\displaystyle l_1}$ in L, which is not in K, i.e, which is not a linear combination of ${\displaystyle k_1,k_2,k_3}$ and ${\displaystyle k_4}$.
So, the set ${\displaystyle \{k_1,k_2,k_3,k_4,l_1\}}$ s a linear independent et of vectors in V. Thus the dimensionality of V is at least 5!
In fact, it is possible for the span of ${\displaystyle \{k_1,k_2,k_3,k_4,l_1\}}$ to be the entire vector space V - so that the minimum number of basis vectors must be 5
Just as an example, let V be ${\displaystyle {\mathbb{R}}^5}$ and let K and V consists of vectors of the forms
${\displaystyle \left(\begin{array}{c}\alpha \\ \beta \\ \gamma \\ \delta \\ 0\end{array}\right)}$ and ${\displaystyle \left(\begin{array}{c}\mu \\ \nu \\ \lambda \\ 0\\ \varphi \end{array}\right)}$
It is easy to see that the vectors
${\displaystyle \left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right)\text{and}\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right)}$
form a basis of K Append the vector ${\displaystyle \left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right)}$ , and you will get a basis for the entire vector space,