From the textbook Morita Equivalence and continuous trace C*algebra:
So I can show that the map $a \mapsto(\pi(a) h \mid h)$ for $h \in \mathcal{H}$ with $\|h\|=1$ is a state if our C* algebra is unital. The main part is to show that the map is norm 1. We can show it is equal to or less than 1, and we can plug in the identity to get that it is in fact 1. But if our C*algebra is non-unital, is it still possible to show this map is a state?
For the argument you are looking at, you don't really need to show that $$\tag1a\longmapsto\langle \pi(a)h,h\rangle$$ is a state; all that is needed is that it is a positive linear functional, because then it can be normalized to be a state. So the question is how to know that it is positive. Depending on what definition of positive you are using, this will require different paths. If you already know that $a\geq0$ if and only if $a=b^*b$ for some $b$, then $$\langle\pi(b^*b)h,h\rangle=\|\pi(b)h\|^2\geq0.$$ If instead the definition is that $a\geq0$ if and only if $a^*=a$ and $\sigma(a)\subset[0,\infty)$ then a bit more work is required. First one immediately gets that if $a\geq0$ then $\pi(a)\geq0$. Then one needs to show that for a selfadjoint operator $T$ one has that its numerical range, that is the set $$ N(T)=\{\langle Th,h\rangle: \ \|h\|=1\}, $$ is the closed convex hull of $\sigma(a)$. It follows that if $N(T)\subset [0,\infty)$ and $T$ is normal, then $T\geq0$. This argument is more sophisticated because (I think) it requires the Spectral Theorem.
I'll finish with some comments about the question of showing that the map in question is in fact a state. I don't think there is an elementary argument in the non-unital case.
Note that the authors are implicitly assuming that $\pi$ is non-degenerate, for otherwise you could have something like $A=\mathbb C$, $H=\mathbb C^2$ and $$ \pi(a)=\begin{bmatrix} a&0\\0&0\end{bmatrix},\qquad\qquad h=\begin{bmatrix} 0\\1\end{bmatrix}. $$ Then you have $\langle\pi(a)h,h\rangle=0$ for all $a\in A$ and the map is not a state.
So to avoid the above you have to assume that $\pi$ is non-degenerate, which means that $\pi(A)H$ is dense in $H$.
Here is a false argument that can give you an idea of what one can try. Given $\def\e{\varepsilon}\e>0$ there exists $a\in A$ such that $\|\pi(a)h-h\|<\e$. Then $$ \langle \pi(a)h,h\rangle=\|h\|^2-\langle \pi(a)h-h,h\rangle\geq1-\e. $$ But the gap in the argument is that a priori the element $a$ could have big norm, and we want it to have norm at most $1$. As far as I can tell, this gap cannot be avoided by elementary results. Basically one wants to know that if $H=\overline{\pi(A)H}$, then $(H)_1=\overline{\pi((A)_1)H}$. That is, that the unit ball of $H$ is the closure of the set of elements $\pi(a)h$ with $\|a\|\leq1$. This is usually proven by using an approximate identity. And if one has an approximate identity, then one can use it directly to show that the map in $(1)$ has norm $1$.