Exercise about subspaces of $\mathbb{R}^n$ 
Consider the $\mathbb{R}$-linear space $\mathbb{R}^n$ and $U, W \le \mathbb{R}^n$ with $\dim(U)=\dim(W)=n-1$.  
  Is it true that there exists $V\le \mathbb{R}^n$ such that $U\oplus V = W \oplus V = \mathbb{R}^n$?

Here is my attempt:
I claim that it is true.
Let $B_U=\{u_1,...,u_{n-1}\}$ and $B_W = \{w_1,...,w_{n-1}\}$ be basis of U and W respectively and take $u_n \in \mathbb{R}^n$ such that $B_U \cup \{u_n\}$ is a basis of $\mathbb{R}^n$. 
Then for every choices of $a_1,...,a_n \in \mathbb{R}$, with $a_n \ne 0$, $B_U \cup \{\sum_{i=1}^{n}a_iu_i\}$ is a basis of $\mathbb{R}^n$ and $U \oplus \operatorname{span}(\sum_{i=1}^{n}a_iu_i)=\mathbb{R}^n$. 
If $u_n \notin W$, then $W \oplus \operatorname{span}(u_n)=\mathbb{R}^n$; otherwise consider $u_1+u_n$, if this does not belong to W, then $W \oplus \operatorname{span}(u_1+u_n)=\mathbb{R}^n$. 
If $u_1+u_n \in W$, then $u_1 \in W$, now repeat with $u_2$ instead of $u_1$ and so on.
In the end or we find $u_m+u_n, m\lt n$ such that $W \oplus \operatorname{span}(u_m+u_n)=\mathbb{R}^n$, or $u_1,...,u_{n-1}\in W$ that is $U=W$ and therefore $W \oplus \operatorname{span}(u_n)=\mathbb{R}^n$.
Is this correct? Thanks in advance
 A: An alternative approach is to let $(u_{1},\ldots,u_{n-1})$ be an orthonormal basis of $U$ (which exists by the Gram-Schmidt orthogonalisation procedure) and similarly let $(w_{1},\ldots,w_{n-1})$ be an orthonormal basis of $W$. There exists the $1$-dimensional space $U^{\perp}$ spanned by a vector $u_{n}$ in $\mathbb{R}^{n}$ which is orthogonal to (and linearly independent of) all vectors in $U$, such that $U \bigoplus U^{\perp} = \mathbb{R}^{n}$. Similarly there is the space $W^{\perp}$ spanned in the same way by $w_{n}$. Consider now the space $V$ spanned by $((u_{n} + w_{n})/2)$. As $u_{n}$ is linearly independent of $u_{i}$ for each $i < n$, so is $(u_{n} + w_{n})/2$ hence $U \bigoplus V = \mathbb{R}^{n}$. Analogous reasoning shows that $W \bigoplus V = \mathbb{R}^{n}$. Looking at your proof, it's not immediately clear to me what space is intended to be $V$.
A: Your proof is fine and this is a slight different approach. Note that $U\cup W$ is not the whole space $\mathbb{R}^n$: if $U\cup W$ is a linear space then necessarily $U\subseteq W$ or $W\subseteq U$ which implies that the dimension of $U\cup W$ is $n-1<n$. See also See Union of two vector subspaces not a subspace?
Now take $v\in \mathbb{R}^n\setminus (U\cup W)$  and let $V$ be the $1$-dimensional vector space generated by $v$. Hence $U\cap V=\{0\}$, $W\cap V=\{0\}$ and therefore
$$U \oplus V =W \oplus V =\mathbb{R}^{n}.$$
