Let $\phi: A\to B$ be an isometric linear map between unital C*-algebras $A$ and $B$ such that $\phi(a^*)=\phi(a)^* (a\in A)$ and $\phi(1)=1$. Show that $\phi(A^+) \subset B^+$.
Clearly $A^+ = \{a^*a ; a\in A\}$, and $$\|\phi(a^*a)\|=\|a^*a\|=\|a\|^2 =\|\phi(a^*)\phi(a)\|$$ Because $\phi$ is an isometry, we have $\phi(a^*a)=\phi(a)^*\phi(a)$ which show that $\phi(A^+)\subset B^+$.
I do not use of unitality of $A,B$ , also $\phi(1)=1$. Please check my argument. If it is correct, we can claim for nonunital C*-algebras, the theorem holds. Is it correct? Thanks in advance.
Your argument is not correct. You say that $\|\phi (a^*a)\|=\|\phi (a)^*\phi (a)\|$ implies that $\phi (a^*a)=\phi (a)^*\phi (a) $, which makes no sense.
Also, without the unital condition the statement is trivially false: take $\phi (x)=-x $.
Now, here is an argument using all conditions. Note that $ \phi$ maps selfadjoints to selfadjoints. For a selfadjoint $a $ we have the following property : with $c=\|a\|$,
$$a\in A^+\iff \|a\|\geq\|a-c 1\|.$$
Then, as $c=\|a\|=\|\phi (a)\|$ and $\|a-c 1\|=\|\phi (a-c 1)\|=\|\phi (a)-c 1\|$, $$a\in A^+\implies \|a\|\geq\|a-c 1\|\implies \|\phi (a)\|\geq\|\phi (a)-c 1\|\implies \phi (a)\in B^+. $$
Edit: for completeness, here is a proof of the claim that $$a\in A^+\iff a=a^*\ \text{ and }c\geq\|a-c 1\|,$$ where $c=\|a\|$.
If $a\in A^+$, then $a=a^*$ and $\sigma(a)\subset[0,c]$. Then $a-c1$ is selfadjoint and $\sigma(a-c1)\subset[-c,0]$. So $\|a-c1\|\leq|-c|=c$.
Conversely, if $a\not\in A^+$, then there exists $\lambda>0$ with $-\lambda\in\sigma(a)$. Then $-\lambda-c\in\sigma(a-c1)$, implying that $\lambda+c=|-\lambda-c|\leq\|a-c1\|$. So $c<\|a-c1\|$.