= Solution
Use coordinates $x^0=t$, $x^1=x$ and first compute the mixed <canonical stress-energy tensor> supplied by spacetime translations:
$$
T^\mu{}_{\nu}=\sum_{j=1}^2\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_j)}\partial_\nu\phi_j-\delta^\mu{}_{\nu}\mathcal L.
$$
Define the energy density $\mathcal E=\phi_{1,t}\phi_{2,t}+c^2\phi_{1,x}\phi_{2,x}$ and $\mathcal P=\phi_{1,t}\phi_{2,x}+\phi_{2,t}\phi_{1,x}$. Direct substitution gives
$$
\boxed{T^\mu{}_{\nu}=\begin{pmatrix}\mathcal E&\mathcal P\\-c^2\mathcal P&-\mathcal E\end{pmatrix}.}
$$
The two wave equations imply
$$
\partial_t\mathcal E=c^2\partial_x\mathcal P,\qquad\partial_t\mathcal P=\partial_x\mathcal E.
$$
For example, differentiating $\mathcal E$ and replacing each $\phi_{j,tt}$ by $c^2\phi_{j,xx}$ gives exactly $c^2\partial_x\mathcal P$ by the <product rule>. These are respectively the two component equations $\boxed{\partial_\mu T^\mu{}_{\nu}=0}$, establishing <stress-energy conservation> explicitly.
If the given metric $g^{\mu\nu}=\operatorname{diag}(1,-1)$ is used to raise the second index in these unrescaled coordinates, the result is $T^{\mu\nu}=\begin{pmatrix}\mathcal E&-\mathcal P\\-c^2\mathcal P&\mathcal E\end{pmatrix}$, still conserved. To display the symmetric relativistic form with light speed $c$, choose $X^0=ct$, $X^1=x$ instead. With $\partial_{X^0}=c^{-1}\partial_t$, the corresponding tensor is
$$
\boxed{\mathcal T^{\mu\nu}=\begin{pmatrix}\mathcal E&-c\mathcal P\\-c\mathcal P&\mathcal E\end{pmatrix},\qquad\partial_{X^\mu}\mathcal T^{\mu\nu}=0.}
$$
This states the coordinate convention explicitly rather than setting $c=1$ in some components but not others.
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