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

$$v= v(\in U_i) + \sum_{\large 1\leq j\leq n; \,j\neq i} 0(\in U_j)$$