If $\{u,v,w\}$ is a basis for $V$, then $\{u+v+w,v+w,w\}$ is also a basis: is this proof correct?  Let $u,v,w \in V$ a vector space over a field F such that $u \neq v \neq w$. If $\{ u , v , w \}$ is a basis for $V$, then prove that $\{ u+v+w , v+w , w \}$ is also a basis for $V$.
Proof:
Let $u,v,w \in V$ a vector space over a field $F$ such that $u \neq v \neq w$. Let $\{ u , v , w \}$ be a basis for $V$. Because $\{ u , v , w \}$ is a basis, then $u,v,w$ are linearly independent and $ \langle \{ u , v , w \} \rangle = V$. 
Let $x \in V$ be an arbitrary vector then $x$ can be uniquely expressed as a linear combination of $\{ u , v , w \}$. Let's suppose $x=au+bv+cw$ for some $a,b,c \in F$. 
On the other hand, let us consider $\{ u+v+w , v+w , w \} \subseteq V$.
Then $$ \begin{align*}
\langle \{ u+v+w , v+w , w \} \rangle &= \{d(u+v+w) + e(v+w) + f(w) \mid d,e,f \in F\} \\ &= \{du + (d+e)v +(d+e+f)w \mid d,e,f \in F \} . \end{align*}$$ 
If $x \in V$, then $x=du + (d+e)v +(d+e+f)w$ is another unique representation of $x \in V$ . Then for any arbitrary $x \in V$, we have $d=a$, $d+e=b $and $d+e+f=c \in F$. 
Because $\{ u , v , w \}$ is a basis for $V$, then $\{ u+v+w , v+w , w \}$ must also be a basis for $V$.
 A: It is easily seen that each of your original basis elements are in the span of $\{u+v+w,v+w,w\}$. Since $\{u, v, w\}$ spans $V$, so does $\{u+v+w,v+w,w\}$ (if $A\subseteq {\rm span} \,(V)$, then ${\rm span}(A)\subseteq{\rm span }(V) \thinspace $).
So, we have a set of three vectors that span a space of dimension three. It follows that the set is independent and thus  a basis.
A: In general I find it much harder to show that a set of vectors spans the vector space, than showing a set is independent. If you want to go about your approach and show they span, what you would need to do is take an arbitrary vector in $V$ and write it is a linear combination of your new set. 
An alternative way is to show that they are independent, which turns out to be quite simple. 
Suppose $$c_1(u + v + w) + c_2(v+w) + c_3(w) = 0$$ then we know that $$c_1u + (c_1 + c_2)v + (c_1 + c_2 + c_3) w = 0$$ but $u, v, w$ are independent and so 
$$c_1 = c_1 + c_2 = c_1 + c_2 + c_3 = 0$$
From here its pretty clear that $c_1 = c_2 = c_3 = 0$ which would prove the claim.
EDIT: Note, if we show that the vectors are independent they must also span $V$, because in any vector space of dimension $n$, if we have $n$ independent vectors, they must span the vector space. 
Similarly, if we were able to show that they spanned the space then they would have to be independent since we have $\dim(V)$ of them
A: The proof is essentially correct, but you do have some unnecessary details. Removing redundant information, we can reduce it to the following:
Let $V$ be a vector space over $F$ and $\{u, v, w\}$ a basis for $V$. Any $x \in V$ can be uniquely written $x = au + bv + cw$ for some $a, b, c \in F$. Let $d, e, f \in F$ be the unique choices so that $d = a$, $e = b - a$, and $f = c - b$; then $x$ can be uniquely written as $x = du + (d + e)v + (d + e + f)w$. Rearranging, we have $x = d(u + v + w) + e(v + w) + fw$. Since this is a unique expression for an arbitrary $x \in V$, then $\{u + v + w, v + w, w\}$ is a basis for $V$.
There is also this nice alternate proof:
Let $V$ be a vector space over $F$ and $\{u, v, w\}$ a basis for $V$. Let's start a new basis with $w$. Since $\{u, v, w\}$ is linearly independent, $v + w \notin span\{w\}$ and $u + v + w \notin span\{w, v + w\}$, so $\{w, v + w, u + v + w\}$ is linearly independent. It is also the same length as a basis, so it must be a basis itself.
