Here is the proof that every Hilbert space is refexive:
Let $\varphi\in\mathcal{H^{**}}$ be arbitrary. By Riesz, there is a unique $f_\varphi\in\mathcal{H^*}$ with
$\varphi(f)=\langle\,f,f_\varphi\rangle$ for all $f \in\mathcal{H^*} $.
Using the same notation and theorem, we have
$\hat{y}_{f_\varphi}(f)= f(y_{f_\varphi})=\langle\,y_{f_\varphi},y_f\rangle=\langle\,f,f_\varphi\rangle=\varphi(f)$
This implies $\hat{y}_{f_\varphi}=\varphi$, thus $\mathcal{H}$ reflexive.
I understood all the steps except for the last implication. Basically, we just showed that $2$ functionals from bi-dual space $\mathcal{H^{**}}$ are the same, why would it imply that $\mathcal{H}$ is reflexive? Any explanation would be highly appreciated!