Proving that Span{u, w} = Span{2u − w, w} Proving that Span{u, w} = Span{2u − w, w}, for all $u, w ∈ R^3$.

Let
$u=\begin{pmatrix} a \\ b \\ c \end{pmatrix}$
$w = \begin{pmatrix} d \\ e \\ f \end{pmatrix}$

$x_1, x_2, x_3, x_4 ∈ R$

$$x_1 \begin{pmatrix} a \\ b \\ c \end{pmatrix} + x_2 \begin{pmatrix} d \\ e \\ f \end{pmatrix} = x_3 \begin{pmatrix}2a-d\\ 2b-e\\ 2c-f\end{pmatrix} + x_4\begin{pmatrix}d\\ e\\ f\end{pmatrix}$$

When $x_1=2, x_2=0, x_3=1, x_4=1$, we get:

$\begin{pmatrix} 2a\\ 2b\\ 2c\end{pmatrix}$ = $\begin{pmatrix}2a-d+d\\ 2b-e+e\\ 2c-f+f\end{pmatrix}$

And therefore $\begin{pmatrix} 2a\\ 2b\\ 2c\end{pmatrix}$ = $\begin{pmatrix} 2a\\ 2b\\ 2c\end{pmatrix}$

Is this proof sufficient enough?
For any $x_1,x_2$ you need to find $x_3,x_4$ such that your equation is satisfied. Then you know $span(u,w)\subset span(2u-w,w)$ and still have to show $span(2u-w,w)\subset(u,w)$.
No, this is not a completed proof; you should show that, there exists a bijection between span(u,w) and span(2u - w, w)