Let 4 linearly dependent vectors such that any 3 of them is independence

Notice that $u in L$. Then since $u in L$ and $u-2v in L$ we also have $v in L$ (we can write $v = frac{1}{2}u - frac{1}{2}(u-2v)$). Now since $u+3v+w in L$, we conclude that $w in L$ as well (since $w = (u+3v+w)-u-3v$). And so finally from the fact that $w+x in L$ we have $x in L$ (since $x = (w+x)-w$). Therefore $text{span}(u,v,w,x) subseteq L$.

Conversely, we also have $L subseteq text{span}(u,v,w,x)$ since the vectors used to define $L$ are all in $text{span}(u,v,w,x)$.

Hence $L=text{span}(u,v,w,x)$, so all that remains is to show that the dimension of $text{span}(u,v,w,x)$ is $3$.