Are three matrices linearly independent and form a basis of $M_2(\mathbb R)$?

You're going about it exactly the right way. EDIT: As David Mitra points out, you have to prove that $c_1=c_2=c_3=0$, not just that $c_1+c_2+c_3=0$.

In fact, you can just think of the matrices as being vectors of length 4: $$\begin{pmatrix}a & b \\ c& d\end{pmatrix}\mapsto (a,b,c,d)$$ and use your knowledge about the linear independence of vectors.

It's perfectly fine (except that, if you want to prove independence, you need to show $c_1=c_2=c_3=0$, not that their sum is 0). Next, do the matrix arithmetic on the left hand side:

$$ c_1 \begin {bmatrix} -1&1 \\\\ -1&1 \ \end{bmatrix} + c_2 \begin {bmatrix} 1&1 \\\\ -1&-1 \ \end{bmatrix} + c_3 \begin {bmatrix} -1&1 \\\\ 1&-1 \ \end{bmatrix} = \begin {bmatrix} -c_1+c_2-c_3&c_1+c_2+c_3\\\\ -c_1-c_2+c_3 & c_1-c_2-c_3 \ \end{bmatrix} ={\bf 0}. $$

Since a matrix is the zero matrix if and only if each of its components is 0, you get the system of equations $$\eqalign{ -c_1+c_2-c_3&=0\cr c_1+c_2+c_3&=0 \cr -c_1-c_2+c_3&=0 \cr c_1-c_2-c_3&=0 } $$