In "Numerical Approximation of Partial Differential Equations" the authors Alfio Quarteroni and Alberto Valli state in remark 5.1.1, that the Riesz Representation Theorem suffices to prove the following Theorem, if $A$ is additionally assumed to by symmetric.
Theorem. (Lax-Milgram) Let $(V, (\cdot, \cdot), \| \cdot \|)$ be a (real) Hilbert space and $A \colon V \to V^*$ a linear, strongly positive, bounded operator, that is there exist $\mu, \beta > 0$ such that \begin{equation*} \langle A u, u \rangle \ge \mu \| u \|^2 \quad \text{and} \quad \langle A u, v \rangle \le \beta \| u \| \| v \| \quad \forall u,v \in V, \end{equation*} where $\langle f, v \rangle_{V^* \times V} =: f(v)$ is the dual pairing of $V^*$ and $V$, for $v \in V$ and $f \in V^*$. Then $A$ is bijective.
Here's is my attempt at a proof only using Riesz. As $A$ is symmetric, bounded and strongly positive, $(u, v)_A := \langle A u, v\rangle$ is a inner product on $V$. As $(V, (\cdot, \cdot)_A)$ is a Hilbert space (due to the boundedness and the strong positivity, the induced norms are equivalent to each other), by the Riesz Representation Theorem, there exists a isometric isomorphism $\iota \colon V^* \to V$ such that $\langle f, v \rangle = ( \iota(f), v )_A = \langle A \iota(f), v \rangle$ for all $f \in V^*$ and all $v \in V$. This implies $\langle f - A \iota(f), v \rangle = 0$ for all $f \in V^*$ and all $v \in V$ and hence $f = A\iota f$ for all $f \in V^*$ and hence $A \iota =$id, so $A$ is invertible with $A^{-1} = \iota$.
Is my proof correct?