I was wondering if someone could verify that this proof is correct. I am also open to any critique. Thanks!
Theorem: Suppose that $V$ is finite dimensional with $\dim{V}=n$. Prove that there exist one-dimensional subspaces $U_1,...,U_n$ such that: $$V = U_1\oplus\cdots\oplus U_n.$$ Proof: We want to show that there exist one dimensional subspaces $U_j$ such that $V = U_1+\cdots+U_n$ and $\sum_{j=1}^{n}\dim{U_j}=\dim{V}$. Suppose $V$ is finite dimensional over $\mathbb{F}$ and $\dim{V}=n$. Because $V$ is finite dimensional, it has a basis $B = (b_1,...,b_n)$. Let $U_j = \text{span}(b_j)$. Because only one (linearly indepedent) vector spans each $U_j$, we must have $\dim{U_j}=1$ and: $$U_1+\cdots+U_n = \{c_1b_1,\ b_1\in B, c_1\in\mathbb{F}\} + \cdots+\{c_nb_n,\ b_n\in B,\ c_n\in\mathbb{F}\}$$ which is equivalent to $\text{span}(B) = V$. Additionally, because each subspace is one dimensional: $$\sum_{j=1}^n\dim{U_j} = \sum_{j=1}^{n}1 = n = \dim{V}.$$ So we must have $$V = U_1\oplus\cdots\oplus U_n.$$