Your assumption is that $\pi$ is faithful. Being a $*$-homomorphism, this implies that $\pi$ is injective, because
$$
\pi(X)=0\ \implies\ \pi(X^*X)=\pi(X)^*\pi(X)=0\ \implies\ X^*X=0\ \implies\ X=0.
$$
Now we need to use that the image of $\pi$ is closed and so it is a C$^*$-algebra (I added a proof at the end).
We showed that $\pi$ is injective, so bijective onto its image. Then $\pi^{-1}$ is a $*$-homomorphism, thus contractive. Then, for any $X\in\mathcal A$,
$$
\|X\|=\|\pi^{-1}\pi(X)\|\leq\|\pi(X)\|\leq\|X\|.
$$
================================
Proof that a homomorphic image of a C$^*$-algebra is closed (this is how Kadison-Ringrose prove it in Theorem 4.1.9 of their first volume).
Let $Y\in\overline{\pi(\mathcal A)}$. We want to show that $Y\in\pi(\mathcal A)$.
We have that $Y=\lim\pi(A_n)$ for some sequence $\{A_n\}$ in $\mathcal A$. As $Y^*=\lim\pi(A_n^*)$, the real and imaginary parts of $Y$ are limits of the corresponding real and imaginary parts of the $A_n$. So we may assume, without loss of generality, that $Y$ is selfadjoint.
Also, by choosing a subsequence if necessary, we may assume that $\|\pi(A_{n+1})-\pi(A_n)\|<2^{-n}$ for all $n$. In particular, $\sigma(\pi(A_{n+1}-A_n))\subset[-2^{-n},2^{-n}]$.
Choose continuous functions $f_n$ with the property that $f_n(t)=t$ when $t\in[-2^{-n},2^{-n}]$ and $|f_n(t)|\leq2^{-n}$ for all $n$. Let
$$
A=A_1+\sum_nf_n(A_{n+1}-A_n)\in\mathcal A
$$
(the series converges because $\|f_n(A_{n+1}-A_n)\|<2^{-n}$ for all $n$).
The key point is that, since $f_n$ is the identity on the spectrum of $\pi(A_{n+1}-A_n)$, it is the identity when evaluated on the operator. Then, as $\pi$ is a continuous $*$-homomorphism,
$$
\pi(A)=\lim_m\pi(A_1)+\sum_{n=1}^m\pi(f_n(A_{n+1}-A_n))\\
=\lim_m\pi(A_1)+\sum_{n=1}^mf_n(\pi(A_{n+1}-A_n))\\
=\lim_m\pi(A_1)+\sum_{n=1}^m\pi(A_{n+1}-A_n)\\
=\lim_m\pi(A_1)+\sum_{n=1}^m\pi(A_{n+1})-\pi(A_n)\\
=\lim_m\pi(A_m)=Y.
$$
The easiest way to show that $\phi$ is isometric goes as follows: Using that $\lVert a \rVert$ equals its spectral radius for self-adjoint $a$, one sees that the norm on $A$ is uniquely determined. Now, define a norm
$$
\rho(a) := \lVert \phi(a)\rVert \qquad (a \in A).
$$
Then the norm $\rho$ makes $A$ into a C*-algebra. Therefore, $\rho = \lVert \cdot \lVert$. It follows that $\phi$ is isometric.
Best Answer
That would be II.2.2.4, the Gelfand Transform, a few theorems back. If you are new to C$^*$-algebras, reading Blackadar (with almost no proofs and a general point of view) is probably not the best idea.
Answer to the edit: as Blackadar says, the C$^*$-identity lets you restrict the problem to $A,B$ abelian. So you have maps $$ C_0(\hat A)\xrightarrow{\ \ \ \ \ } A\xrightarrow{\ \ \phi\ \ } B\xrightarrow{\ \ \ \ \ } C_0(\hat B), $$ where the three maps are injective ($\phi$ by hypothesis and the other two by II.2.2.4). Now when restricted to algebras of continuous functions you can use II.2.2.5/II.2.2.7: say $\psi:C(X)\to C(Y)$ is an injective homomorphism. Then $\breve\psi:Y\to X$ is surjective. Now, for $f\in C(X)$, you have (I do the compact case to avoid a few epsilons) $$ \|f\|=|f(x_1)|=|f(\breve\psi (y_1))|=|(\psi f)(y_1)|\leq \|\psi f\|. $$ Also, $$ \|\psi f\|=|(\psi f)(y_0)|=|f(\breve\psi(y_0))|\leq \|f\|. $$ Thus $\|f\|=\|\psi f\|$.