as title
Here's a sketch of a proof.
Let U and V be the subspaces. If U≠V then neither of U and V can contain the other. So there exist vectors $\displaystyle u\in U\setminus V$ and $\displaystyle v\in V\setminus U$. Then $\displaystyle w_1\mathrel{\mathop=^{\mathrm{d{}ef}}}u+v$ is in neither U nor V. If U=V then take $\displaystyle w_1$ to be any vector not in U.
Now proceed by an inductive construction. Let $\displaystyle U_1$ be the subspace spanned by U and $\displaystyle w_1$, and let $\displaystyle V_1$ be the subspace spanned by V and $\displaystyle w_1$. These spaces have the same dimension as each other. If they are not the whole space then we can repeat the construction in the previous paragraph to get a vector $\displaystyle w_2$ that is in neither $\displaystyle U_1$ nor $\displaystyle V_1$. Continue doing this, getting a sequence of vectors $\displaystyle w_1, w_2, w_3,\ldots$. On the assumption that the ambient spaces is finite-dimensional, this construction will stop after a finite number of steps. The subspace spanned by $\displaystyle w_1, w_2, w_3,\ldots$ will then be complementary for both U and V.