jmc

If $(x,y)$ lies on the circle, so does $(x,-y),$ $(-x,-y),$ and $(-x,-y),$ (which all give the same value of $|x|+|y|$), so we can assume that $x\ge 0$ and $y\ge 0.$

Then $|x|+|y|=x+y.$ Squaring, we get $$(x+y{)}^2=x^2+2xy+y^2=1+2xy.$$Note that $(x-y{)}^2\ge 0.$ Expanding, we get $x^2-2xy+y^2\ge 0,$ so $2xy\le x^2+y^2=1.$ Hence, $$1+2xy\le 2,$$which means $x+y\le \sqrt{2}.$ Equality occurs when $x=y=\frac{1}{\sqrt{2}},$ so the maximum value of $|x|+|y|$ is $\boxed{\sqrt{2}}.$