First using Brouwer
Let $w\in \mathcal{V}_k$, necessarily $\zeta(w)$ is a bounded function since $\phi_k $'s are also bounded. The Lax-Milgram lemma implies there is a unique function $\widehat{w}\in \mathcal{V}_k$ such that \begin{align} \int_\Omega\nabla\widehat{w}\nabla\phi +\zeta(w)\widehat{w}\phi -f\psi dx=0\quad\text{for all}\quad\phi\in \mathcal{V}_{k}. \end{align}
The Poincar'{e}--Friedrichs inequality yields \begin{equation} \int_\Omega|\nabla\widehat{w}|^2dx + \int_\Omega \zeta(w )\widehat{w}^2\, d x\leq \|f\|_{L^{2}(\Omega)}\,\|\widehat{w}\|_{L^{2}(\Omega)}\leq C\|f\|_{L^{2}(\Omega)}\, \|\widehat{w}\|_{H_0^1(\Omega )} \end{equation}
Thus, letting $R=C\,\|f\|_{L^{2}(\Omega)}$, since $\varphi\geq0 $ we obtain the following estimates \begin{align}\label{eq:boundedmapT} &\|\widehat{w}\|_{H_0^1(\Omega )}\leq R \quad\text{ and }\quad \int_\Omega \zeta(w )\widehat{w}^2\,dx\leq R^2. \end{align} We let $ \mathcal{B}_R=\big\{ w\in \mathcal{V}_k: \|w\|_{H_0^1(\Omega)} \leq R\big\}$, be the closed ball in $\mathcal{V}_k$ of radius $R$ centered at the origin. Clearly, the mapping $T:\mathcal{V}_k\to \mathcal{B}_R$ with $Tw=\widehat{w}$ is well defined.
It remains to prove that $T$ is a continuous mapping. Indeed, let $\{w_n\}$ be a sequence in $\mathcal{V}_k$ with $w_n= \lambda_{1,n}\phi_1+\cdots+ \lambda_{k,n}\phi_k$ converging in $\mathcal{V}_k$ to a function $w= \lambda_1\phi_1+\cdots+ \lambda_k\phi_k$; i.e., $\lambda_{\ell,n}\xrightarrow{n\to\infty } \lambda_\ell$, $\ell=1,2,\cdots,k$. By continuity we have $ \varphi(w_n)\xrightarrow{n\to \infty}\varphi(w)$ almost everywhere. In addition, the convergence in $L^2(\Omega)$ also holds, since the continuity gives $\sup_{n\geq 0} \|\varphi(w_n) \|_{L^\infty(\Omega)} <\infty$ because $\sup_{n\geq 0}\|w_n\|_{L^\infty(\Omega) } <\infty$.
On the other side, in virtue of the first estimate above the sequence $\{Tw_n\}$ is bounded in finite dimensional space $\mathcal{V}_k$ and thus converges $\mathcal{V}_k$ up to a subsequence to some $w_*\in \mathcal{V}_k$. Altogether, it follows that, for all $\phi\in \mathcal{V}_k\subset L^\infty(\Omega)$ % \begin{align*} (f,\phi)= \lim_{n\to \infty} \int_\Omega\nabla\widehat{w}_n\nabla\phi +\zeta(w_n)\widehat{w}_n\phi dx= \int_\Omega\nabla\widehat{w}\nabla\phi +\zeta(w)w_* \phi. \end{align*}
The uniqueness of $\widehat{w}$ entails $w_*=\widehat{w}=Tw$ and hence the whole sequence $\{Tw_n\}$ converges in $Tw$ in $\mathcal V_k$, which gives the continuity of $T$.
Therefore, by the Brouwer fixed-point theorem, $T$ has a fixed point $v_k\in \mathcal V_k$, i.e., $v_k=Tv_k$ which clearly satisfies the announced relation.
An alternative.
The given problem is equivalent to the minimization problem \begin{align} \mathcal{J}(v_k)= \min_{v\in \mathcal{V}_k} \mathcal{J}(v)\quad \text{with}\quad \mathcal{J}(v):= \frac12 \int_\Omega|\nabla v|^2 dx + \int_\Omega G(v)d x + \int_\Omega fvd x \end{align} and we define the function $G(v)= \int_0^v\zeta(\tau)\tau d \tau $. Note that $G$ is non-negative since $\zeta(\tau)\geq 0$ and that $\mathcal J$ is continuous on $\mathcal V_k$. Using the Poincaré-Friedrichs inequality we find that $\mathcal{J}(v)\to \infty$, as $\|v\|_{L^2(\Omega)} \to \infty$ and $v\in \mathcal{V}_k$. Which implies the existence of a minimizer $v_k\in \mathcal{V}_k$ of \mathcal{J}, since we are in finite dimension space.