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Suppose $U$ is a subspace of $V$ such that $V/U$ is finite-dimensional. Prove that there exists a subspace $W$ of $V$ such that $\dim W = \dim V/U$ and $V=U\oplus W$.

Here's what I've got so far: $V/U$ is finite-dimensional so it must have a finite basis $\{v_1 + U, \cdots, v_n+ U\}$. Choose any representative element of each $v_i + U$ and call it $w_i$ (I'm not sure if this is allowed). Then let $W = \operatorname{span}(w_1, \dots, w_n)$. Then $\dim W = n = \dim V/U$. Let $v\in V$. Then $$v+U \in V/U \implies v+U = a_1(w_1 + U) + \cdots + a_n(w_n + U) \\ v+U = (a_1w_1 + \cdots + a_nw_n) + U \\ v-(a_1w_1 + \cdots + a_nw_n) \in U$$ Which tells us that $v-(a_1w_1 + \cdots + a_nw_n) = u$ for some $u\in U$. Denote $a_1w_1 + \cdots + a_nw_n$ by $w$. Note that $w\in W$. Thus $v = u + w$ for $U$ and $W$.

I'm not sure how to show this is a unique decomposition of $v$, though.

Am I on the right track here? If so, how do I show that $u$ and $w$ are unique?

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    Yes, you're on the right track. To show that $w$ is unique just quotient by $U$. Then it follows that $u = v - w$ is unique. – 2017-02-24

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Suppose that $v=u'+w'$ with $u'\in U$ and $w'\in W$. Then $$w'-w=u-u'\in U$$ so $$(w'-w)+U=U\ .$$ Clearly $w'-w\in W$, write $$w'-w=a_1w_1+\cdots+a_nw_n\ .$$ Then $$a_1(w_1+U)+\cdots+a_n(w_n+U)=(w'-w)+U=U$$ and by independence $a_1=\cdots=a_n=0$. So $w'=w$ and $u'=u$.

BTW to conclude that $\dim W=n$ you need to show that $w_1,\ldots,w_n$ are linearly independent. The argument for this is pretty much the same as the one I have just given.