Find a subspace $W$ of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U \oplus W$
Let $W=\{(0,w,z,0)\in\mathbb{F}^4:w,z\in\mathbb{F}\}$, then we check that such $W$ is the desired subspace, as the following two steps.
- Given $(a,b,c,d)\in\mathbb{F}^4$, it is easy to decompose the vector as below. \begin{align} (a,b,c,d) &=(a,b-a+a,c-d+d,d)\\ &=(a,a,d,d)+(0,b-a,c-d,0), \end{align} where $(a,a,d,d)\in U$ and $(0,b-a,c-d,0)\in W$. Hence $\mathbb{F}^4=U+W$.
- If $(e,f,g,h)\in U\cap W$, then we have $e=f$, $g=h$, $e=0$, and $h=0$. So $$(e,f,g,h)=(0,0,0,0)$$ and hence $U\cap W=\{(0,0,0,0)\}$.
$ U $ is a subspace spanned by the linearly independent set $ S = \{ (1, 1, 0, 0), (0, 0, 1, 1) \} $. Therefore, it suffices to pick a subspace $ W $ which is spanned by two vectors such that their adjoinment to $ S $ would not disturb its linear independence. In other words, we need to extend $ S $ to a basis of $ \mathbb{F}^4 $.
It is easy to see that $ (1, 0, 0, 0), (0, 0, 0, 1) \notin U $. Now, we check if the set $ S' $ formed by adjoining these vectors to $ S $ is linearly independent. The standard method is to row reduce the matrix whose columns are elements of $ S $, but a more direct approach works here. Let $ s_i $ denote the elements of $ S' $:
$$ c_1 s_1 + c_2 s_2 + c_3 s_3 + c_4 s_4 = ( c_1 + c_3, c_1, c_2, c_2 + c_4) $$
For the left hand side to equal zero, it is then clear that we must have $ c_i = 0 $ for all coefficients, establishing linear independence of $ S' $. Therefore, $ S' $ is a basis of $ \mathbb{F}^4 $ (by the dimension theorem), and we may take $ W = \textrm{span} \{(1, 0, 0, 0), (0, 0, 0, 1) \} = \{ (x, 0, 0, y) : x, y \in \mathbb{F} \} $.