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Without loss of generality, we may assume $a\ge b\ge c$. The triangle inequality and $a+b+c=1$ imply that $a<b+c=1-a$, i.e. $a<\frac{1}{2}$. Since $b\le a$, it follows that $\sqrt{a^2+b^2}\le \sqrt{2a^2}=a\sqrt{2}<\frac{\sqrt{2}}{2}$. Since $c\le b$, it follows that $b^2+c^2\le b^2+bc<b^2+bc+\frac{c^2}{4}={\left(b+\frac{c}{2}\right)}^2$, i.e. $\sqrt{b^2+c^2}<b+\frac{c}{2}$.

Analogously we conclude $\sqrt{a^2+c^2}<a+\frac{c}{2}$. Summing the obtained inequalities we get $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}<\frac{\sqrt{2}}{2}+\left(b+\frac{c}{2}\right)+\left(a+\frac{c}{2}\right)=1+\frac{\sqrt{2}}{2}.$$