I am trying to prove that given $$g:I^{n+1}\subset \mathbb{R}^{n+1}\rightarrow \mathbb{R}^n$$ continuous, $(x_0,y_0)\in I^{n+1}$ and a sequence of functions defined iteratively as for each $f:\mathbb{R}\rightarrow \mathbb{R}^n$, $$ f_0(x)=y_0\\ f_{n+1}(x)=\vec{y_0}+\int_{x_0}^xg(s,f_n(s))ds $$ is continuous on the compact k-cell $I^{n}$. I am not seeing an obvious way to bound the difference $$ ||\int_{x_0}^xg(s,f_n(s))ds-\int_{x_0}^yg(s,f_n(s))ds|| $$ in terms of $||x-y||$. If it were a single variable integral this would be easy, but I am having a hard time trying to pluck out $||x-y||$ terms from anywhere.
Some context: I have (hopefully) shown uniform boundedness of the $f_n$ by arguing, for $L=||I^{n}||$ and $||g||_\infty\leq M$, $$ ||f_n(x)||_\infty=||\vec{y_0}+ \int_{x_0}^xg(s,f_n(s))\mathrm ds||_\infty\\ \stackrel{\text{triangle ineq.}}{\leq}||y_0||_\infty+ ||\int_{x_0}^xg(s,f_n(s))\mathrm ds||_\infty\\ \stackrel{\text{max of integrand times max volume}}{\leq} ||y_0||_\infty+ML $$ a constant.
This is in the context of proving a variation of the existence theorems for ODE's using Arzela-Ascoli.
edit: addressing comment: I may have left out some context because I was confused myself. The $g$ in this problem arises in the context of the system of differential equations $$ f'(x)=g(x,f(x))\\ f(x_0)=y_0 $$ for a continuous $g$ defined on an open interval $\mathcal{O}\subset \mathbb{R}^n\times \mathbb{R}\rightarrow \mathbb{R}^n$. I took $I^{n+1}$ to be some compact product of closed rectangles living inside $\mathcal{O}$, analogous to the picard theorem in 1 dimension.
The $f_n$ are approximating the differentiable $f:\mathbb{R}^n\rightarrow \mathbb{R}$.
As long as you know that $g(s,f_n(s))$ always makes sense (i.e. $(s,f(s))$ lies in the domain of the function $g$), and $g$ is bounded by $M$ on this set, then it's not too bad to get what you're after. Assume for the moment that $y \le x$. Then $$ \int_{x_0}^x g(s,f_n(s)) ds - \int_{x_0}^y g(s,f_n(s)) ds = \int_{y}^x g(s,f_n(s)) ds $$ and so $$ \left\Vert \int_{x_0}^x g(s,f_n(s)) ds - \int_{x_0}^y g(s,f_n(s)) ds \right\Vert \le \int_y^x \Vert g(s,f_n(s)) \Vert ds \le \int_y^x M ds = M |y-x|. $$ A similar argument proves the bound when $y \ge x$. Thus we have something better than continuity: we have Lipschitz continuity.