Let $I=\int_{C}\sqrt{x^2+y^2}ds$. So far, I have only managed to prove that $I\ge\sqrt{2}-{1\over2}$, the proof of which is below.
Let $c(t)$ be a unit speed parametrization of $C$ such that $c(0)=(0,1)$ and $c(L)=(1,0)$, where $L$ is the length of $c$. Let $f(x,y)=\sqrt{x^2+y^2}$. Notice that $f(c(t))=\|c(t)\|$ and $\|c'(t)\|=1$ since $c$ is unit speed.
Then, $I=\int_{C}fds=\int_{0}^Lf(c(t))\|c'(t)\|dt=\int_{0}^L\|c(t)\|ds$. Since $c$ is unit speed, $-1\le{d\over{dt}}\|c(t)\|dt\le1$. Also, $\|c(0)\|=\|c(L)\|=1$. So, if $L\ge2$, then $\int_{0}^{1}\|c(t)\|dt\ge{1\over2}$. Similarly, $\int_{L-1}^{L}\|c(t)\|dt\ge{1\over2}$, so $\int_{0}^{L}\|c(t)\|dt\ge{1}$, meaning we now only need consider $L<2$.
If $L<2$, then $\int_{0}^{L\over2}\|c(t)\|dt\ge{L\over{2}}(1-{L\over2})+{L^2\over8}\le\int_{L\over2}^{L}\|c(t)\|dt$, so $\int_{0}^{L}\|c(t)\|dt)\ge{L-{L^2\over4}}$.
Finally, since the distance between $(0,1)$ and $(1,0)$ is $\sqrt2$, $L\ge\sqrt2$. Since $L-{L^2\over4}$ is increasing over the interval $[\sqrt2,2]$, this implies $I\ge\sqrt2-{{\sqrt2}^2\over4}=\sqrt2-{1\over2}$.
This bound is somewhat close to $1$ at approximately $0.914...$, however I have been unable to find any way to proceed with improving upon my current inequality. How can I prove the desired result?
Suitably interpreted (for example by Riemann sum) with $A=(0,1)$ and $B=(1,0)$, $$\int_A^B \sqrt{x^2+y^2}\cdot \sqrt{dx^2+dy^2}\geq \int_A^B |xdx|+|ydy|$$$$\geq \frac 12+\frac 12=1,$$ where the first inequality follows from C-S, and the last minimum is achieved if both $x$ and $y$ are monotone.