Let the four vectors k_1,k_2,k_3 and 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 l_1 in L, which is not in K, i.e, which is not a linear combination of k_1,k_2,k_3 and k_4.
So, the set {k_1,k_2,k_3,k_4,l_1} is a linear independent set of vectors in V. Thus the dimensionality of V is at least 5!
In fact, it is possible for the span of {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 RR^5 and let K and V consists of vectors of the forms
((alpha),(beta),(gamma),(delta),(0)) and ((mu),(nu),(lambda),(0),(phi))
It is easy to see that the vectors
((1),(0),(0),(0),(0)),((0),(1),(0),(0),(0)),((0),(0),(1),(0),(0))and ((0),(0),(0),(0),(0))
form a basis of K. Append the vector ((0),(0),(0),(0),(0) ), and you will get a basis for the entire vector space,
