Could anyone check my proof?
Statement: Prove that four vectors of the three dimensional Euclidean space are always linearly dependant.
Proof: A group of vectors are linear dependent if their determinant is zero. Suppose we have three linearly independent vectors,$\textbf{u},\textbf{v},\textbf{w}$, in $\Bbb{R}^3$. Now consider the subspace $(a,b,c,0)$ of $\Bbb{R}^4$ which is isomorphic to $\Bbb{R}^3$. We map the three vectors to that subspace. If we add another vector $\textbf{x}$ to $(a,b,c,0)$, which is the same as adding another vector to $\Bbb{R}^3$, we see that the determinant of the four vectors is equal to zero. Therefore, four vectors in three dimensional Euclidean space are always linearly dependent.
Proof(Attempt again):Suppose we have four linear independent vectors,$\textbf{u},\textbf{v},\textbf{w}$ and $\textbf{x}$ in $\Bbb{R}^3$. Then $a_{1}\textbf{u}+a_{2}\textbf{v}+a_{3}\textbf{w}+a_{4}\textbf{x}=\textbf{0}$ only when $a_{1}=a_{2}=a_{3}=a_{4}=0$. However, \begin{bmatrix}u_{x} & v_{x} & w_{x} & x_{x}\\u_{y} & v_{y} & w_{y} & x_{y}\\u_{z} & v_{z} & w_{z} & x_{z}\end{bmatrix} becomes \begin{bmatrix}u_{x} & v_{x} & w_{x} & x_{x}\\0& v'_{y} & w'_{y} & x'_{y}\\0 & 0 & w'_{z} & x'_{z}\end{bmatrix} by carrying out row operations.
Assuming that the vectors are not zero vectors this shows that there is another non-zero solution for $a_4$ other than zero which is $\frac{-w'_{z}}{x'_{z}}$. Therefore any four vectors in $\Bbb{R}^3$ is linearly dependent.
Talking about determinants does not make sense too much since one needs a square matrix to start with and four $3-dimensional$ vectors will not give a square matrix.
However, you can work with a similar argument. If you have four linearly independent vectors, that means any three of them are also linearly independent, say $a,b,c$.
Then if you take the vector on your basis you didn't include for those three vectors, it can be obtained by $a,b,c$.
I will leave the details of the proof/arguments to you.