Linear dependency proof. Let $x_1, x_2, x_3 \in \mathbb{R}^n$ be linearly independent vectors in $n$-dimensional space. Define
$$u=x_1+x_2, v=x_1+x_3, w = x_2 + x_3.$$
Are the vectors $$u,v,w$ linearly independent?

What I did is:
We know that
$$a\cdot x_1 + b\cdot x_2 + c\cdot x_3 = 0 \iff a = b = c =0$$
Now , for
\begin{align*}
l\cdot u + m\cdot v + n \cdot w &= 0 \\ 
l\cdot (x_1+x_2) + m\cdot (x_1+x_3) + n\cdot (x_2+x_3) &= 0\\
lx_1 + lx_2 + mx_1 + mx_3 + nx_2 + nx_3 &= 0 \\
(l+m)\cdot x_1 + (l+n)\cdot x_2 + (m+n)\cdot x_3 &= 0\\
\end{align*}
so $$(l+m) = (l+n) = (m+n) = 0$$
The reason I am thinking is,
since assume $l+m = 0 \rightarrow l = -m$, also $l+n = 0 \rightarrow l = -n$, at a time $l$ cannot be equal to $m$ and $n$, the only possibility is that $m==n$, Then the other tern $(m+n)=0$, says that $m=-n$, so no number other than $0$, is equal to its own negative number ,
Thus it can be said that , $$l=m=n=0,$$ thus the vectors $(u,v, w)$ are linearly independent.
My question is what I did here is correct or not, and is there any other intuitive way to prove this.
 A: Your solution is correct and probably the easiest way of proving that the vectors are independent.

There are just some minor issues with the expressions you use. Nothing tragic, just a little awkward. You say that

at a time l cannot be equal to m and n, the only possibility is that m==n

This sounds strange and also sounds as if you are making a proof by contradiction. In fact, all you are doing is applying the transitivity property which claims that if $A=B$ and $B=C$ then $A=C$. In your case, you can simply state that $-m=l$ and $l=-n$, therefore, $-m=-n$ and $m=n$.
From that, you say "no other number other than $0$ is equal to its own negative number" whih, again, is true, but unnecessary complication. You can simply say that $$m+n=0,$$
and because $m=n$, this means $m+m=0$ which means $2\cdot m=0$ which means $m=\frac02=0$.

Oh and one more mathematical strangeness you wrote. Instead of writing
$$"\Rightarrow \text{  iff}"$$
you should just write $$"\iff"$$
I edited your question accordingly.
A: Another way to do it is to show that one can recover $x_1,x_2,x_3$ from linear combinations of $u,v,w$ e.g. $$2x_3 = w + v - u.$$ And since the span of $u,v,w$ contains $x_1,x_2,x_3$ it follows that $u,v,w$ must be also span the (3 dimensional) subspace generated by $x_1,x_2,x_3$ and thus linearly independent.
