Dimensions of vector subspaces in a direct sum are additive $V = U_1\oplus U_2~\oplus~...~ \oplus~ U_n~(\dim V < ∞)$ $\implies \dim V = \dim U_1 + \dim U_2 + ... + \dim U_n.$ [Using the result if $B_i$ is a basis of $U_i$ then $\cup_{i=1}^n B_i$ is a basis of $V$]
Then it suffices to show $U_i\cap U_j-\{0\}=\emptyset$ for $i\ne j.$ If not, let $v\in U_i\cap U_j-\{0\}.$ Then
\begin{align*}
v=&0\,(\in U_1)+0\,(\in U_2)\,+\ldots+0\,(\in U_{i-1})+v\,(\in U_{i})+0\,(\in U_{i+1})+\ldots\\
& +\,0\,(\in U_j)+\ldots+0\,(\in U_{n})\\
=&0\,(\in U_1)+0\,(\in U_2)+\ldots+0\,(\in U_i)+\ldots+0\,(\in U_{j-1})+\,v(\in U_{j})\\
& +\,0\,(\in U_{j+1})+\ldots+0\,(\in U_{n}).
\end{align*}
Hence $v$ fails to have a unique linear sum of elements of $U_i's.$ Hence etc ...
Am I right?

Yes, you're correct. 
Were you second guessing yourself? If so, no need to:
You're argument is "spot on".
If you'd like to save yourself a little space, and work, you can write your sum as:
$$ \dim V = \sum_{i = 1}^n \dim U_i$$
"...If not, let $v\in U_i\cap U_j-\{0\}.$ Then 
$$v= v(\in U_i) + \sum_{\large 1\leq j\leq n; \,j\neq i}  0(\in U_j)$$
