From Morris, A. O., Linear Algebra, an introduction (2nd edition, Van Nostrand, 1989) he gives the following as not being an inner product.
$(u,v)=x_1y_1-x_2y_1-x_1y_2+2x_2y_2$, where $u=(x_1,x_2),v=(y_1,y_2)$
My working below shows otherwise. Where do I go wrong?
Let $w=(z_1,z_2)$
$u+w=(x_1+z_1,x_2+z_2)$
$(u+w,v)=<(x_1+z_1,x_2+z_2),(y_1,y_2)>$
$=(x_1+z_1)y_1-(x_2+z_2)y_1-(x_1+z_1)y_2+2(x_2+z_2)y_2$
$=x_1y_1+z_1y_1-x_2y_1-z_2y_1-x_1y_2-z_1y_2+2x_2y_2+2z_2y_2$
$=x_1y_1-x_2y_1-x_1y_2+2x_2y_2+z_1y_1-z_2y_1-z_1y_2+2z_2y_2$
$=(u,v)+(w,v)$.
Similarly $(\alpha u,v)=\alpha(u,v)$
$(u,v)=(v,u)$ can easily be verified, and
$(u,u)=x_1y_1-x_2x_1-x_1x_2+2x_2x_2$
$=x_1^2-2x_1x_2+x_2^2+x_2^2$
$(x_1-x_2)^2+x_2^2 >0$ if $u \neq 0$.
That product can be written as $$ (u, v) = u^{t} A v $$ where $$ A = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}. $$ Since $A$ is symmetric, this is a perfectly good inner product as you've already shown. It's even positive-definite since $$ (u,u) = x_1^2 - 2x_1 x_2 + 2x_2^2 = (x_1-x_2)^2 + x_2^2 $$ is semipositive, and zero if and only if $x_1 = x_2 = 0$. Morris must have meant to write something else, but good for you to find the error!