I'm not sure if I understand the problem correctly; I'm asking for some feedback on my proof attempt. (Note: this is a modified version of an exercise in Bosch's 'Lineare Algebra' and the translation for 'generating set' is 'Erzeugendensystem'.)
$\Rightarrow$
Let $V=\oplus^{n}_{i=1}V_i$ be a direct sum decomposition of a $k$-vector space into subspaces $V_i\subset V$ and $\forall i\in\{1,...,n\}$ a family $(x_{ij})_{j\in J_i}$ of elements $x_{ij}\in V_{i}$. Show that the elements $x_{ij}, i\in \{1,...,n\}, j\in J_i$ build a generating set of $V$ iff $\forall i\in \{1,...,n\}$ the elements $x_{ij}, j\in J_i$ build a generating set of $V_i$.
In the textbook, the $j\in J$ index usually denotes Kronecker's delta.
My attempt: the direct sum of subspaces can be written in this case as: $\prod^{n}_{i}V_i=(x_1, ...,x_n)$. It means that each vector $x\in V$ can be written as a linear combination: $(\alpha_1 x_1,...,\alpha_n x_n)$ ($\alpha_i$ are scalars). But that implies that also each $x_{ij}$ can be written as: $\alpha_i(0,0,0...,x_i, 0,0,0)$, with $x_{ij}$ defined as $x_i \Leftrightarrow i=j$ and $0 \Leftrightarrow i\neq j$. This implies that for every $i$, $x_i$ is a generating set of $V_i$.
Is my understanding of the problem going in the right direction? I'd be grateful for any hints.
Sorry, but you seem not to have understood the problem.
You have, for each $i$, a family $$ \mathcal{A}_i=\{x_{i1},x_{i2},\dots,x_{im_i}\} $$ of vectors in $V_i$.
Suppose the union of these families spans $V$ and you want to see whether $\mathcal{A}_i$ spans $V_i$ (for all $i$). Consider $v\in V_k$. Then, by assumption, you can write $$ v=\sum_{ij}\alpha_{ij}x_{ij}= \sum_{j}\alpha_{kj}x_{kj}+ \sum_{\substack{ij\\i\ne k}}\alpha_{ij}x_{ij} $$ In particular $$ \sum_{\substack{ij\\i\ne k}}\alpha_{ij}x_{ij} = v-\sum_{j}\alpha_{kj}x_{kj} $$ belongs to $V_k$, but also to $$ V_1\oplus\dots\oplus V_{k-1}\oplus V_{k+1}\oplus V_n $$ What can you conclude?
What about the converse? Hint: a vector $v\in V$ can be written as $v=v_1+\dots+v_n$, with $v_i\in V_i$.