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
$$\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{a^2 + c^2} < \frac{\sqrt{2}}{2} + \left(b + \frac{c}{2}\right) + \left(a + \frac{c}{2}\right) = 1 + \frac{\sqrt{2}}{2}.$$