Part 1   \(\cal{D}\) spans \(W\):   If \(\Vect{w}\) is an arbitrary vector in \(W\), we need to show that it can be expressed as a linear combination of vectors in \(\cal{D}\). Indeed, we know that there exist \(\Vect{u}\) in \(U\), and \(\Vect{v}\) in \(\Vect{V}\) such that

\[\Vect{w} = \Vect{u} + \Vect{v}\]

But then we know that \(\Vect{u}\) is a linear combination of vectors in \(\cal{B}\):

\[\Vect{u} = s_1\Vect{b}_1 + \cdots + s_m\Vect{b}_m\]

and, similarly, that \(\Vect{v}\) is a linear combination of vectors in \(\cal{C}\):

\[\Vect{v} = t_1\Vect{c}_1 + \cdots + t_n\Vect{c}_n\]

Therefore

\[\Vect{w} = s_1\Vect{b}_1 + \cdots + s_m\Vect{b}_m\ +\ t_1\Vect{c}_1 + \cdots + t_n\Vect{c}_n\]

is a linear combination of vectors in \(\cal{D}\).

Part 2   \(\cal{D}\) is linearly independent:   Consider the situation

\[\Vect{0} = (s_1\Vect{b}_1 + \cdots + s_m\Vect{b}_m)\ +\ (t_1\Vect{c}_1 + \cdots + t_n\Vect{c}_n) = \Vect{0} + \Vect{0}\]

Then \((s_1\Vect{b}_1 + \cdots + s_m\Vect{b}_m)=\Vect{0}\) because and \((t_1\Vect{c}_1 + \cdots + t_n\Vect{c}_n)=\Vect{0}\). But then

\[s_1, \dots , s_m = 0\]

because \(\cal{B}\) is linearly independent, and

\[t_1, \dots , t_n = 0\]

because \(\cal{C}\) is linearly independent.

Part 3   the dimension formula:   follows by counting the vectors in the basis \(\cal{D}\) of \(W\).