I'm trying to prove the following statement...
Let $f:[a,b] \times \mathbb{R} \to \mathbb{R}$ a bounded and continuous function, $t_{0} \in [a,b]$, $x_{0} \in \mathbb{R}$, $r>0$ and $$B= \{ x \in C([a,b]) : ||x-x_{0}||_{\infty} \leq r \}.$$ The application $\phi:B \to C([a,b])$ defined by $\phi(x)=g$ with $$g(t)= x_{0} + \int_{t_{0}}^{t} f(s,x(s)) ds$$ is continuous for $||\cdot||_{\infty}$.
I know that if I had $f:[a,b] \times [c,d] \to \mathbb{R}$ then I could prove the continuity of $\phi$ using the uniform continuity of $f$. But in this case I don't know how to do it.
I have a small hint: To prove the continuity in $y \in B$ use the uniform continuity of $f$ in the compact set $[a,b] \times K$, where $K=y([a,b]) + [-1,1]$ but I don't know how yo use it. Thanks in advance.
By the way, I have proved that $\phi(B) \subseteq B$ and that $g$ is bounded and equicontinuous, in case it is useful.
Since you are working in a normed space you could work with sequences and show $$ f_n\rightarrow f \Rightarrow \phi(f_n) \rightarrow\phi(f) $$
with respect to the $||.||_\infty$ norm. Alternatively, you can try to prove $\varepsilon, \delta$ continuity, which in normed spaces is equivalent. In this case, this seems to be simpler to me. You have ($x_0$ cancels out) $$ ||\phi(x)-\phi(y)||_\infty = \sup_t\left|\int_{t_0}^t \left( f(s, x(s) -f(s, y(s)) \right) ds \right| \le{ \sup_t\int_{t_0}^t|f(s, x(s)) -f(s, y(s))|ds}$$
Since $f$ is continuous, it is uniformly continuous on compact sets, e.g. on the square $[a-b]\times[0,r+||x_0||_\infty] $ (you say you know $\phi(B)\subset B$).
So if $\varepsilon >0$ there is $\delta$ such $||x-y||_\infty < \delta$ implies for $x,y\in B$ that $\sup_s |f(s, x(s)) -f(s, y(s))|<\varepsilon$ (the first argument is the same).
Then the last displayed formula implies (on $B$!) $$ ||\phi(x)-\phi(y)||_\infty \le \sup_t \int_{t_0}^t ds\,\varepsilon \le (b-a) \varepsilon $$