I'm trying to solve this question here:
Let $V = F_{n-1}\left [ x \right ]$ over some field F (i.e. V is the vector space of all polynomials with degree smaller or equal to n-1), and $x_{1}, x_{2}, ..., x_{n} \in F$ be n different scalars. Assume $A_{1}, ... , A_{k}$ are disjoint sets, such as $\left \{ x_{1},...,x_{n} \right \} = \sqcup _{i=1}^{k} A_{i}$. Define $V_{i} = \left \{ p(x)\in V \mid \forall x_{j} \notin A_{i}, p(x_{j}) = 0 \right \}$.
We need to prove that $V = \oplus _{i=1}^{k} V_{i}$.
I managed to prove this in the case in which $k=n, A_{i} = \left \{ x_{i} \right \}$, but I'm having trouble how to use this in order to prove the general case.
The nice thing about this question is you can be extremely concrete. Taking the $k=n$, $A_i=\{x_i\}$ case, we see that we have a basis $e_i(x)=\prod_{i\neq j} \frac{(x-x_j)}{x_i-x_j}$ of $V$. (Clearly each $e_i\in V_i$, $e_i(x_i)=1$, and moreover if $\sum_j a_j e_j(x)=0$, then evaluating at $x_i$, $a_i=\sum a_j e_j(x_i)=0$. Since we have $n$ linearly independent vectors in an $n$-dimensional space, it is a basis.)
Now we apply this to the general case by expressing everything in this basis. First, we have $V_i=\bigoplus_{x_j\in A_i} F e_j(x)$. Indeed, clearly each $e_j\in V_i$ if $x_j\in A_i$, and conversely any $p(x)\in V_i$ can be written $p(x)=\sum a_i e_i(x)$, and satisfies $0=p(x_j)=a_j$ for $x_j\notin A_i$, hence must be in the span of the $e_j$ with $x_j\in A_i$.
Finally, note that the span of the $V_i$ contains the span of the basis, hence $\sum V_i=V$, and since $V_i$ and $V_j$ contain distinct basis elements when $i\neq j$ (due to the disjointness of $A_i$ and $A_j$), $V_i\cap V_j=\emptyset$. The result follows.