Showing that ${1 + i,−1 + i}$ is a basis for the vector space $\mathbb C$ over $\mathbb R$ I feel my justification for this is weak, and I'm seeking for improvement.
$$Span((1+i), (-1+i)) = a(1+i) + b(-1+i)$$
I have two conclusions from this:
1.
$$a(1+i) + b(-1+i) = a + ai -b + bi$$
$$a(1+i) + b(-1+i) = (a - b) + i (a +b)$$
And, other than just saying by setting $a$ and $b$ to any real number, any vector in $\mathbb C$ can be created.
2.
$a(1+i)$ is a vector that resembles the image of $y = x$ in that it can extend to any point in that direction.
$b(-1+i)$ is a vector that resembles the image of $y=-x$ in that it can extend to any point in that direction. 
Since these two vectors are orthogonal, any linear combination of the two can form any vector in $\mathbb C$.

Is this a valid proof? Or is there a better way, or proper way, to show this? Besides, the span of any two linearly independent vectors in $\mathbb C$ should be able to span all of $\mathbb C$, anyway. An example would be $(1+i)$ and $(3+2i)$.
 A: By definition, the numbers $1$ and $i$ are a basis the vector space $\mathbb C$ over $\mathbb R$, so this is a two-dimensional $\mathbb R \text{-}$vector space. 
In general, a nonzero vector $\vec{w}$ is linearly independent from a nonzero $\vec{v}$ iff for any $\lambda$, $\; \vec{w} \ne \lambda \vec{v}$. In particular, the vectors 
$\quad +1 + i$
$\quad −1 + i$
are linearly independent, and therefore form a basis.
You can also span all vectors with these two complex numbers. To see this, we can easily get back the 'canonical' basis in the span,
$(1 + i) + (−1 + i) = 2i,$ so $i = (0,1)$ is a $\checkmark$,
$(1 + i) - (i) = 1,\; \; \;\;\;\;\;\;\;$ so $1 = (1,0)$ is a $\checkmark$,
and since the span in closed under any (recursive) linear combinations you can form, you can now 'get to' any vector (number) in $\mathbb C$.
A: Since the linear combination is
$(a-b)+(a+b)i$,
all you need to do
is to state that,
for any $c+di$,
you can make
$(a-b)+(a+b)i
=c+di$
by setting
$a=(c+d)/2$
and
$b = (d-c)/2$.
Nothing more.
A: Your 1) is a good basis for a proof, but the conclusions can be better formulated.
You shown that a linear combination   of the two given complex numbers  $u_1=1+i$ and $u_2=-1+i$ has the form $(a-b)+i(a+b)$ with $a,b \in \mathbb{R}$. From this you can conclude that, for any complex number $z=x+iy$ we can use 
$$
a=\frac{x+y}{2} \qquad b=\frac{y-x}{2}
$$
to represent the number $z$ as a linear combination of the two numbers $u_1,u_2$. This also prove that the dimension of the vector space $ \mathbb{C}$ over $\mathbb{R}$ is $2$.
Your second ''proof'' is more a geometric intuition than a proof but it is a bit confusing because, as you noted, the vectors of  basis can also be not orthogonal.
A: you need to add a proof for that $\forall x,y\;,x=a-b,y=a+b$
it can easily be proved: for given $x,y$ i can construct $2$ equations for $a$, \begin{cases}a=x+b\\a=y-b\end{cases}so now we can find $b$, $x+b=y-b\implies\frac{y-x}{2}=b$. after i have $b$ i can find $a$ by putting $b$ to one of the equations: $a=x+\frac{y-x}{2}=\frac{2x}{2}+\frac{y-x}{2}=\frac{y+x}{2}$. after adding this your proof is done.

another way is to show that $(1+i)\ne k(-1+i)$
it can be done easily by contradiction: if the above equation can be solved than $\begin{cases}1=-k\\i=ik\end{cases}$ but if $i=ik$ then $1=k$ and if $1=-k$ then $-1=k$ and we get $1=k=-1$. by this contradiction we conclude that $Span((1+i), (-1+i))$ is $\Bbb C$.

both of those ways are valid
Edit
another way is to show that $\arg(1+i)-\arg(-1+i)\not\equiv 0\pmod\pi$
