Line Integral gives no work done?

Here's how physicists often do it:

$$int_C mathbf{F}cdot dmathbf{r} = int_C (F_x hat{i} + F_y hat{y})cdot (hat{i},dx + hat{j},dy) = int_C F_x,dx + F_y,dy$$

On the first part $x$ goes $0 to 1$ and $y=0$ is constant so $dy = 0$ and hence $$int_{C_1} F_x,dx + F_y,dy = int_{x=0}^{x=1} -y,dx = 0.$$

Similarly, on the second part $y$ goes $0 to 1$ and $x=1$ is constant so $dx = 0$ and hence $$int_{C_2} F_x,dx + F_y,dy = int_{y=0}^{y=1} x,dy = int_{y=0}^{y=1} dy=1.$$

Your error is in the second integral. Along the path from $(0, 0)$ to $(1, 0)$ we can take as parameterization $x= t$ (from $0$ to $1$), $y= 0$ for all $t$. So $F(x,y)= langle-y, xrangle= langle 0,t rangle$ and the vector differential is $langle dt, 0 rangle$ so the integral is $$int_0^1 langle 0, t rangle cdot langle dt, 0 rangle = int_0^t 0= 0$$ That is what you correctly have.

Along the path from $(1, 0)$ to $(1, 1)$ we can take as parameterization $x= 1$ for all $t$, $y= t$ (from $0$ to $1$). So $F(x,y)= langle -y, x rangle= langle -t, 1 rangle$ (not $langle-t, 0 rangle $ because $x= 1$) and the vector differential is $langle 0, dy rangle$ so the integral is $$int_0^1 langle -t, 1 rangle cdot langle 0, dy rangle = int_0^1 dy= 1$$

So the complete integral is 1. Again, your error is that on the second line, from $(1, 0)$ to $(1, 1)$ as $x$ is always $1$, not $0$.

Your second integral is wrong. Note that $x=1$ on this portion of the path.