In a proof of the Riesz representation theorem, I saw the following argument concerning the isometry ($H$ is a Hilbert space and $H^{*}$ its topological dual) :
Take $\varphi\in H^{*}$. There exists $y\in H$ such that $\varphi(x) = \langle x,y\rangle$ $\forall x\in H$. Then by Cauchy-Schwartz we have $\lvert \varphi(x)\rvert = \lvert\langle x,y\rangle\rvert\leq\lVert x\rVert_{H}\lVert y\rVert_{H}$ which implies $\lVert \varphi\rVert\leq\lVert y\rVert_{H} $
Is this follows by the fact that $\lVert \varphi\rVert = sup_{x\neq 0_{H}}\frac{\lvert\varphi(x)\rvert}{\lVert x\rVert_{H}}$ ? Which is the norm I have naturally considered when trying to prove this inequality.
If yes, is the result only true for this specific norm on $H^{*}$ ? Thank you
Yes, the inequality $\lVert \varphi \rVert_{H^*} \leq \lVert y \rVert_H$ follows from the definition of the (canonical) dual norm: $$ \lVert \varphi \rVert_{H^*} = \sup \left\{ \frac{|\varphi(x)|}{\lVert x \rVert } \; : \; x \in X, x \neq 0 \right\}$$ and from the Cauchy-Schwartz inequality, as you have noticed.
This holds for this specific norm: the dual norm on $H^*$ (which can also be shown to be induced by an inner product, thus turning $H^*$ into an Hilbert space).
Actually, under these same hypotheses, you have the equality $\lVert \varphi \rVert_{H^*} = \lVert y \rVert_H$, implying that the map $H \to H^*$ implicitely given by the theorem is an isometry (this is usually the second part of the statement of the Riesz representation Theorem).