How to show that $(I - F)(U)$ is open, when $U$ is open and $F$ is a contraction? 
Let $U$ be an open subset of a Banach space $E$ and let $F:U \to E$ be a contraction. Show that $(I - F)(U)$ is open.

This is an exercise on page 9 of Fixed point theory and applications, Ravi Agarwal et al. $I$ is an identity map. I don't know where to start.
Added: A map $F:X \to X$ is said to be a contraction, iff for all $x,y \in X$,there exists $L<1$ such that:
$$d(F(x),F(y)) \leq Ld(x,y)$$ 
 A: Edit: I have elaborated on my approach below.


*

*First show that $I - F$ is a bijection on $E$.

*Then show that $I - F$ has a continuous inverse by showing that if $(I - F)x_n = y_n$, $(I - F)x = y$, and $\|y_n - y\| < \delta$, then there exists $C > 0$ such that $\|x - x_n\| < C \delta$.
Proof. Let $y \in E$. Then there exists $x \in E$ such that $(I - F)x = y$ if and only if $x = F(x) + y$. If we define $G(x) = F(x) = y$, then this is equivalent to $x = G(x)$; but $G$ is also a contraction since $F$ is, so $x$ exists and is unique by the contraction mapping principle; hence $I - F$ is a bijection; so $(I - F)^{-1}$ exists. Now let $(I - F)x_n = y_n, (I - F)x = y$, and $\|y_n - y\| < \delta$. Then
$$(1 - L) \|x_n - x\| \le \|x_n - x\| - \|F(x_n) - F(x)\| \le \|(I - F)x_n - (I - F)x\| < \delta$$
in which case $\|x_n - x\| < (1 - L)^{-1} \delta$; thus $(I - F)^{-1}$ is continuous. $(I - F)^{-1}$ is continuous, so $(I - F)$ is an open map.
A: Let $\hat{y} = \hat{x}-F(\hat{x})$, with $\hat{x} \in U$. Define $\phi_y(x) = F(x)+y$, and note that $\hat{x}$ is a fixed point of $\phi_\hat{y}$. It is easy to see that $\phi_y$ is Lipschitz (in $x$) with constant $\lambda \in (0,1)$ (the Lipschitz constant of $F$).
Now let $x_n(y) = \phi_y^n(\hat{x})$, and note that for $y$ close to $\hat{y}$, $x_n(y)$ will be close to $\hat{x}$, and hence $x_n(y) \in U$ (so it is defined in a neighborhood of $\hat{x}$). The usual line of argument shows that $\|x_{n+1}(y)-x_n(y)\| \le \lambda^n \|\phi_y(\hat{x}) -\hat{x}\|$, which gives the estimate $\|x_n(y)-\hat{x} \| \leq \frac{1}{1-\lambda} \|\phi_y(\hat{x}) -\hat{x}\|$. 
Suppose $B(\hat{x},\delta) \subset U$, for some $\delta >0$. Now choose $\epsilon>0$ such that if $y \in B(\hat{y},\epsilon)$, then $\frac{1}{1-\lambda} \|\phi_y(\hat{x}) -\hat{x}\| < \frac{\delta}{2}$. Then the above estimate shows that $x(y) = \lim_n x_n(y)$ satisfies $x \in B(\hat{x},\delta)$. Since $x(y)$ is a fixed point of $\phi_y$, we have $y = x(y) - F(x(y)) \in (I-F)(U)$. Hence $B(\hat{y},\epsilon) \subset (I-F)(U)$, and so $(I-F)(U)$ is open.
(As an aside, I find the book Kantorovich & Akilov, "Functional analysis" to be an excellent reference for these sorts of extimates.)
