If $U \leq \mathbb{R}^4$, $\dim(U) =3$ and $\langle(0,0,0,1)\rangle \cap U = \{0\}$ then $U = \langle(1,0,0,0), (0,1,0,0), (0,0,1,0)\rangle$ 
Suppose $U \leq \mathbb{R}^4$, $\operatorname{dim}(U)=3$ and $\langle(0,0,0,1)\rangle \cap U =  \{0\}$. Is it then true that $U = \langle(1,0,0,0), (0,1,0,0), (0,0,1,0)\rangle$?

I say yes. Here's my reasoning:
Put the three basis vectors of $U$ into RRE form. That is consider the matrix with the three basis vectors as rows. The only RRE form that doesn't contain $(0,0,0,1)$ is the matrix $\begin{bmatrix}1&0 &0 & 0\\0&1&0 &0 \\ 0 &0 &1 &0\end{bmatrix}$. This proves the claim.
Is this correct?
 A: No, take $U=\text{span}\{(1,0,0,0),(0,1,0,0),(0,0,1,1)\}$ which has dimension $3$ since $\{(1,0,0,0),(0,1,0,0),(0,0,1,1)\}$ is clearly linearly independent. If $U$ were equal to $\text{span}\{(1,0,0,0),(0,1,0,0),(0,0,1,0)\}$, then $U$ would have $(0,0,1,1)-(0,0,1,0)=(0,0,0,1)$ as an element and would therefore equal all of $\mathbb{R}^4$, which is impossible. Furthermore, $U$ does not contain a nonzero multiple of $(0,0,0,1)$ since that would also imply that $U=\mathbb{R}^4$ by similar reasoning.
The issue with your argument is that the RRE of the matrix whose rows are basis vectors may not be the matrix you listed. With the example I listed before, the RRE form of the matrix and will be
$$\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&1\end{pmatrix}.$$
If, alternatively, you made the basis vectors the columns of the matrix and put it into RRE form, the reasoning would fail for a different reason. That is, the column space of a matrix is not always the same as the column space of the RRE form version of that matrix. The two column spaces will be $\textit{isomorphic}$ since they will have the same dimension, but they may not be $\textit{equal}$ as subsets of the ambient vector space. For a simple example of this phenomenon, take the matrix
$$A=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$$
whose column space is the $y$ axis in $\mathbb{R}^2$. The RRE form of $A$ is $\begin{pmatrix}1&0\\ 0&0\end{pmatrix}$ whose column space is the $x$-axis.
