5. Given $x, y, z \in(0,1)$ and $x+y+z=2$, then the range of $x y+y z+z x$ is ( ).
(A) $\left(0, \frac{4}{3}\right]$
(B) $\left(1, \frac{4}{3}\right]$
(C) $\left(1, \frac{4}{3}\right)$
(D) $\left(\frac{4}{3}, 2\right)$
(5) B Hint: Let $S=x y+y z+z x$, then
$$
(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 S .
$$

Since $x, y, z \in(0,1)$, we have $S1$.
Also,
$$
\begin{array}{c}
S^{2} \leqslant\left(x^{2}+y^{2}+z^{2}\right)^{2}=(4-2 S)^{2}, \\
(3 S-4)(S-4) \geqslant 0,
\end{array}
$$

Solving this, we get $S \leqslant \frac{4}{3}$ or $S \geqslant 4$, but since $S<3$, we have $S \leqslant \frac{4}{3}$, and when $x=y=z=\frac{2}{3}$, $S=\frac{4}{3}$.
In summary, $1<S \leqslant \frac{4}{3}$. Therefore, the answer is B.