It is not clear to me why you assume that $u\mapsto\zeta(u)u$ is bounded - this does not follow from the assumption that its derivative is bounded on every compact subset.

But let us assume this. Then the finite-dimensional equation is of the form
$$Av+F(v)=0$$
where $A$ is linear and positive definite, and $F$ is continuous and bounded. In particular, $A^{-1}$ exists, and the equation thus is equivalent to
$$v=-A^{-1}F(v)\text.$$
The range of the map $G=-A^{-1}F$ is contained in some ball. In particular, $G$ maps this ball into itself, and so Brouwer's fixed point theorem implies that $G$ has a fixed point which thus is a solution of the finite-dimensional equation.