$\dim(U)=\dim(U_1)+\dots+\dim(U_n)\implies$ every $x\in U$ has a unique representation $x=u_1+\dots+u_n$, with $u_i\in U_i$ 
Let $U$ be a finite dimensional vector space, and $U=U_1+U_2+\dots+U_n$ with $U_i$ subspaces, show that if $\dim(U)=\dim(U_1)+\dots+\dim(U_n)$ then every $x\in U$ has a unique representation $x=u_1+\cdots+u_n$, with $u_i\in U_i$.

for $n=2$ assume the converse, $x=u_1+u_2=u'_1+u'_2$, then $0=(u_1-u'_1)+(u_2-u'_2)\in U_1\cap U_2\Rightarrow(u_2-u'_2)\in U_1\cap U_2$, so by dimension formula $\dim(U_1+U_2)=\dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2)$, and $U_1\cap U_2$ is nonempty, but what if $n\ge2$ then the dimension formula looks complicated by inclusion exclusion etc. 
 A: For all $i$, let $B_i$ be a basis of $U_i$. Since $U = \sum_i U_i$, you have that
$\cup_i B_i =B$ generates $U$. Moreover, by your assumption, you have
$$|\cup_i B_i| \le \sum_i |B_i| = \sum_i \dim U_i = \dim U \le |\cup_i B_i|$$
hence equality holds. This means that $B$ is a basis of $U$, and that all the $B_i$s are pairwise disjoint. This concludes the proof.
A: Using induction: assume that 
$$
\dim U = \dim U_1 + \cdots + \dim U_n \implies {  \rm every  }\,x\ldots
$$
then, assuming
$$\dim U = \dim U_1 + \cdots + \dim U_{n+1}$$
and defining $V =U_1 + \cdots + U_n $: $U = V + U_{n+1}$.
$$
\dim U \le \dim V + \dim U_{n+1} \le \dim U_1 + \cdots + \dim U_{n} + \dim U_{n+1} = \dim U
$$
so every inequality is an equality and in particular:
$$ \dim V = \dim U_1 + \cdots + \dim U_{n} $$
(so every $v\in V$ writes itself uniquely as a sum $u_1 + \cdots + u_n$ where $u_i\in U_i$) and 
$$
\dim U = \dim V +  \dim U_{n+1} 
$$so every $u\in U$ writes itself uniquely as $u = v + u_{n+1}$, $v\in V$ and 
$u_{n+1}\in U_{n+1}$.

combining both, you get the result.
