If I understand correctly then for an operator $\mathcal{A}$ defined on a Hilbert space $\mathcal{H}$, $\langle \mathcal{A}x,x\rangle\geq 0$ does not necessarily imply that $\mathcal{A}$ is Hermitian: $\mathcal{A} = \mathcal{A}^\ast$. See for instance Is a positive operator symmetric?
However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?
Lemma 1: $\langle Tx,x\rangle = 0$ for every $x\in \mathcal{H}$ implies that $T \equiv 0$.
Proof: We show that $\langle Tx,y\rangle = 0$ for every pair $x,y\in \mathcal{H}.$ Indeed
\begin{align*} 0 = \langle T(x+y),x+y\rangle - \langle T(x-y),x-y\rangle & = 2\langle Tx,y\rangle+2\langle Ty,x\rangle. \end{align*} This implies that $$\langle Tx,y\rangle = -\langle Ty,x\rangle.$$
Exchanging $x$ with $ix$ yields
$$0 = i\langle Tx,y\rangle -i\langle Ty,x\rangle$$ why $$\langle Tx,y\rangle = \langle Ty,x\rangle$$ all in all we find that $\langle Tx,y\rangle = \pm \langle Ty,x\rangle$ which implies that both are $0$.
Proof that $\langle \mathcal{A}x,x\rangle\geq 0$ implies that $\mathcal{A}^\ast = \mathcal{A}$:
We have that
$$\mathbb{R}\ni\langle \mathcal{A}x,x\rangle = \langle x,\mathcal{A}^\ast x\rangle = \overline{\langle \mathcal{A}^\ast x,x\rangle } = \langle \mathcal{A}^\ast x, x\rangle\Rightarrow \langle (\mathcal{A}-\mathcal{A}^\ast)x,x\rangle = 0$$ for every $x$ thus by the lemma $\mathcal{A}-\mathcal{A}^\ast\equiv 0$.
In the case complex scalars $\langle Tx,x \rangle=0$ for all $x$ implies $T=0$ and $\langle Tx,x \rangle \geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $\mathbb R^{2}$ is a counter example.