# Consequence of the Lax-Milgram theorem

I have the following problem: Let H be a $\mathbb{K}$-Hilbertspace, with finite dimensional subspaces $H_n\subset H, n\in \mathbb{N}$, so that $H_n\subset H_{n+1} \forall n\in \mathbb{N}$ and $\bigcup_{n=1}^\infty H_n$ is dense in H. Also $x'\in H'$ and $a:H \times H\rightarrow \mathbb{K}$ is a continuous, coercive Sesquilinear form.

i) Show: For every sequence $(x_n)_{n\in \mathbb{N}}$ in H is true: $$x_n\rightarrow x\Leftrightarrow x_n\rightharpoonup x \cap lim_{n\rightarrow \infty} a(x_n,x_n)=a(x,x)$$ for $n\rightarrow \infty$

ii) Show: $\forall n\in \mathbb{N}$ there is exactly one $x_n\in H_n$ so: $$a(y,x_n)=x'(y)\forall y\in H_n$$

iii) $(x_n)_{n\in\mathbb{N}}$ is the in ii) constructed sequence and $x\in H$ is the unique solution of $a(y,x)=x'(y)\forall y\in H$. Show: $x_n\rightarrow x$ for $n\rightarrow \infty$

Let $A\in L(H)$ be defined like in the Lax-Milgram theorem and $H$ is a Hilbertspace and $J$ is the isometry of the Riesz representation theorem. For $u'\in H'$, $u:=A^{-1}J^{-1}u'$ is the unique solution to $$a(v,u)=u'(v) \forall v\in H$$
It is also true: $$||u||_H\le\frac{1}{\alpha} ||u'||_{H'}$$
Since $a$ is a continuous, coercive Sesquilinear form I could use this theorem if I'm able to show that every $H_n$ is a Hilbertspace as well. Is it right to say that, since the $H_n$ are subspaces they are also Hilbertspaces? If that is the case then ii) should be proofed, right?