Let \begin{align} B_1\,&=\,\big(u_{11},\dots,u_{1n_1} \big),\dots,B_p\,=\,\big(u_{p1},\dots,u_{pn_p} \big) \end{align} be disjoint linearly independent sets such that $$B\,=\,B_1\cup\cdots\cup B_p $$ is linearly independent. Then $$\langle B\rangle \,=\,\langle B_1\rangle\oplus\cdots\oplus\langle B_p\rangle. $$ My proof :
Since $B_1,\cdots,B_p$ are disjoint and linearly independent, we have $B$ is a basis of $\langle B\rangle$. Hence, for each $u\in\langle B\rangle$, there exists uniquely coefficients $a_{11},\cdots,a_{1n_1},\cdots,a_{p1},\cdots,a_{pn_p}$ such that \begin{align} u\,=\,\big(\underbrace{ a_{11}u_{11}+\cdots+a_{1n_1}u_{1n_1}}_{\in\langle B_1 \rangle} \big)+\cdots+\big(\underbrace{ a_{p1}u_{p1}+\cdots+a_{pn_p}u_{pn_p}}_{\in\langle B_p\rangle } \big) \end{align} Therefore, $$\langle B\rangle \,=\,\langle B_1\rangle\oplus\cdots\oplus\langle B_p\rangle. $$ May anyone tell me if my proof is valid ? My thanks a lot.
Answer BEFORE requiring $B\,=\,B_1\cup\cdots\cup B_p$ to be linearly independent:
You can't assert that $B$ is a basis of $\langle B \rangle$. For example, in the $\mathbb R$-vector space $\mathbb R$, consider $B_1=\{1\},B_2=\{2\}$. Obviously each $B_i$ is linearly independent since they are singletons and also they are disjoint, but $\langle B_1\rangle\cap\langle B_2\rangle=\langle\{1\}\rangle\cap\langle\{2\}\rangle=\mathbb R\cap \mathbb R=\mathbb R\not=\{0\}$.
Edit:
If $B\,=\,B_1\cup\cdots\cup B_p$ is linearly independent your proof seems correct to me. I just would add, to make it clearer, that the uniqueness of the coefficients $a_{11},\ldots,a_{1n_1},\ldots,a_{p1},\ldots,a_{pn_p}$ also implies that the expression of $u$ as a sum of vectors of $\langle B_i \rangle$ is unique, from where it follows that the sum is direct.