Why is every positive linear map between $C^*$-algebras bounded?

We know that every positive linear functional on a $C^*$-algebra is bounded.

How can we prove every positive linear map between $C^*$-algebras is bounded?

Let $\phi:A\longrightarrow B$ be a positive linear map between two $C^*$-algebras. We will show that $\phi$ is automatically bounded. We can assume that $B$ is unital without loss of generality, but this does not change anything. The problem is with $A$. Here is a proof I learned in Blackadar's book which works whether $A$ is unital or not, for positive linear functionals like for positive linear maps in general. We will eventually use the partial order on self-adjoint elements $x\leq y$ if $y-x$ is positive. By assumption, $x\leq y$ implies $\phi(x)\leq \phi(y)$. Note that for $z$ self-adjoint in $B$ unital, we have $\|z\|\leq M$ if and only if $-M1\leq z\leq M1$.

1) Every $x\in A$ can be written $x=h+ik$ with $h,k$ self-adjoint, $\|h\|\leq \|x\|$ and $\|k\|\leq \|x\|$. Indeed, set $h:=\frac{x+x^*}{2}$ and $k:=\frac{x-x^*}{2i}$. So it suffices to prove that $\phi$ is bounded on self-adjoint elements.

2) Every self-adjoint $x=x^*\in A$ can be written $x=x_+-x_-$ with $x_+,x_-$ positive and $\|x_+\|\leq \|x\|$, $\|x_-\|\leq \|x\|$. Indeed, by functional calculus, take $f_+(t)=\max(t,0)=\frac{|t|+t}{2}$ and $f_-(t)=\max(-t,0)=\frac{|t|-t}{2}$. Then set $x_+:=f_+(x)$ and $x_-:=f_-(x)$. So it suffices to prove that $\phi$ is bounded on positive elements.

3) Assume for a contradiction that $\phi$ is not bounded on positive elements and take $x_n$ a sequence of positive elements such that $\|x_n\|=1$ and $\|\phi(x_n)\|\geq 4^n$. Now set $x:=\sum_{n\geq 1}2^{-n}x_n\in A$. For every $n\geq 1$, $x\geq 2^{-n}x_n$ whence $\phi(x)\geq \phi(2^{-n}x_n)$ and therefore $\|\phi(x)\|\geq \|2^{-n}\phi(x_n)\|\geq 2^{-n}4^n=2^n$. Contradiction since $\|\phi(x)\|<\infty$.

Note: with more work, we can prove that if $A,B$ are unital and if $\phi:A\longrightarrow B$ is a unital linear map, then $\phi$ is positive if and only if $\|\phi\|=1$.