Continuity of implicit function

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I find a proof of the following theorem in A.N. Kolmogorov and S.V. Fomin's Элементы теории функций и функционального анализа (pp. 492-493 here):

Let $X,Y,Z$ be Banach spaces, $U$ a neighbourhood of point $(x_0,y_0)\in X\times Y$ and $F$ a map of $U$ into $Z$ having the following properties:

  1. $F$ is continuous in point $(x_0,y_0)$.
  2. $F(x_0,y_0)=0$.
  3. Partial [Fréchet] derivative $F'_y(x,y)$ exists in $U$ and is continuous in point $(x_0,y_0)$, and operator $F'_y(x_0,y_0)$ has bounded inverse.

Then [...] there exist $\varepsilon>0,\delta>0$ and a map $$y=f(x)\quad\quad(1)$$defined for $\|x-x_0\|<\delta$ and continuous in point $x_0$ such that every couple $(x,y)$, for which $\|x-x_0\|<\delta$ and $y=f(x)$, verify equation $$F(x,y)=0,\quad\quad(2)$$and, vice versa, every couple satisfying equation $(2)$ and conditions $\|x-x_0\|<\delta$, $\|y-y_0\|\le\varepsilon$ verifies equation $(1)$.

In a following corollary the book says that it is easy to show that if $F$ is continuous in $U$ then $f$ is continuous in a neighbourhood of $x_0$. I find this last statement interesting, but I cannot prove it to myself. Could anybody explain why $f$ is continuous in some neighbourhood of $x_0$ under the hypothesis of the continuity of $F$ in $U$? I heartily thank you!