$τ: Y \to Χ$ is a continuous map and $A: C(X)\to C(Y)$ is defined by $(Af)(y) = f(τ(y))$. How $||A||=1$? The following is from Conway's Functional Analysis :

If $X$ and $Y$ are compact spaces and $τ: Y \to Χ$ is a continuous map, define $A: C(X) \to C(Y)$ by $(Af)(y) = f(τ(y))$. Then $A \in \mathcal{B} (C(X), C(Y))$ and $||A||=1$.

Q.1 - $f$ is continuous because $f \in C(X)$ and $τ$ is continuous by hypothesis, then so is their composition with respect to y meaning that $Af$ is continuous so I couldn't reduce it to the continuity of $A$ and use equivalence of boundedness and continuity; so how $A \in \mathcal{B} (C(X), C(Y))$?
Q.2 - Why $||A||=1$? An attempt solution in here is unclear for both inequalities to reach $||A||=1$.

Showing linearity is not hard, simply check the axioms.
Showing boundedness is easy: since $\tau(Y) \subseteq X$ is a compact set
$$||Af-Ag|| = \max_{y \in Y} |f( \tau (y)) - g( \tau (y))| = \\ = \max_{x \in \tau (Y)} |f(x)-g(x)| \le \max_{x \in X} |f(x)-g(x)| = ||f-g||$$
Hence $||A|| \le 1$.
