Furthermore assume that $F \subseteq \mathbb{R}^{m}$ and is closed and $T \subseteq \mathbb{R}^{n}$ is any subset. By being a contraction on $F$ I mean $\exists \lambda$ 0 < λ < 1 such that $\rvert \phi (x,t) - \phi(y,t) \lvert \leq \lambda \rvert x-y \lvert \forall x,y \in F$ and $t \in T$ then $\exists x(t) \in F$ continuous, such that$\phi(x(t),t)=0\forall t\in T$
It strongly reminds me of the fixed point theorem, then in these terms what it saying is that $x(t)$ is constant for all t (thinking in the point of view of differential equations, there is a $x(t)$ such that $\phi (x,t)$ is its derivative) but I`m needing some help to understand how being a contraction implies that there exists a solution that is constant (is it even the right way to approach?) Can someone link other questions or sites on the internet?
Thanks.
Denoting $\varphi_t := \varphi(\cdot, t) : F \rightarrow F,$ we see that we can apply the Banach-Caccioppoli-Picard Theorem (since $F$ is closed) to obtain the existence of a unique $x_t \in F$ such that $\varphi_t (x_t) = \varphi(x_t, t) = x_t$ for all $t \in T.$ Now let $(t_n)_{n \in \mathbb{N}} \subseteq T$ be a sequence such that $\lim_{n \to \infty} t_n = t \in T.$ We need to show that $\lim_{n \to \infty} x_{t_n} = x_t.$ We have that: $$\vert x_{t_n} - x_t \vert = \vert \varphi(x_{t_n}, t_n) - \varphi(x_t, t) \vert \leq \vert \varphi(x_{t_n}, t_n) - \varphi(x_{t_n}, t) \vert + \vert \varphi(x_{t_n}, t) - \varphi(x_t, t) \vert \leq \\ \varepsilon + \lambda \vert x_{t_n} - x_t \vert$$ for sufficiently large $n$ since $\varphi$ is continuous, so for these sufficiently large values of $n$ we get that $$\vert x_{t_n} - x_t \vert \leq \frac{\varepsilon}{1 - \lambda},$$ so indeed $t \mapsto x_t$ is continuous. I hope this helps. :)
P.S.: As for the existence of a function such as the one you described, I'm afraid it's not the case that it exists in general. Indeed, let $$\varphi(x, t) := \frac{\lambda t}{2} x + 1 - \frac{\lambda t}{2}$$ for some $0 < \lambda < 1$ defined on $[1, 2] \times [0, 1].$ Clearly this is a contraction in the first variable and continuous function and its fixed point for fixed $t$ is $x_t = 1.$ Moreover, this function doesn't have any zeroes (for fixed t or otherwise), so this is why I assumed from the beginning that the statement proved above is actually what you might be looking for.