The map $\phi : H \to H'$ given by $\phi(v) = f_v$, where $f_v(x) = \langle x, v\rangle$, for $x \in H$ is an antilinear bijective isometry. Bijectivity and norm-preservation both follow from the Riesz Representation Theorem, and antilinearity can be easily verified:
$$\phi(\alpha v + \beta w)(x) = f_{\alpha v + \beta w}(x) = \langle x, \alpha v + \beta w \rangle = \overline{\alpha}\langle x, v\rangle + \overline{\beta}\langle x, w\rangle = \overline{\alpha}f_v(x) + \overline{\beta}f_w(x) = \big(\overline{\alpha}\phi(v) + \overline{\beta}\phi(w)\big)(x)$$
Hence, $\phi(\alpha v + \beta w) = \overline{\alpha}\phi(v) + \overline{\beta}\phi(w)$.
Now, as you can show, if two spaces are "antilinearly isometric", then one is complete if and only if the other one is complete.
Indeed, let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $H$. We have $f_n = \phi(x_n)$ for some $x_n \in H$.
The sequence $(x_n)_{n=1}^\infty$ is a Cauchy sequence in $H$:
$$\|x_m - x_n\| = \|\phi(x_m - x_n)\| = \|\phi(x_m) - \phi(x_n)\| = \|f_m -
f_n\| \xrightarrow{m, n \to\infty} 0$$
Since $H$ is complete, $(x_n)_{n=1}^\infty$ converges. Set $x_n \xrightarrow{n\to\infty} x \in H$.
We claim that $f_n \xrightarrow{n\to\infty} \phi(x)$ in $H'$. Indeed:
$$\|\phi(x) - f_n\| = \|\phi(x) - \phi(x_n)\| = \|\phi(x - x_n)\| = \|x - x_n\| \xrightarrow{n\to\infty} 0$$
Hence, $H'$ is complete.
Your attempt is fine, but for the direction $\overline{A} = H \implies A^{\perp} = \{0\}$ I would suggest not using the generic $h$ at all, apply the argument directly to $h = x \in A^{\perp}$.
Since $A$ is dense, there is a sequence $(a_n)_{n \in \mathbb{N}}$ in $A$ with $a_n \to x$. Then, since $x \in A^{\perp}$, we have
$$\lVert x\rVert^2 = \langle x, x\rangle = \lim_{n \to \infty} \langle a_n,x \rangle = \lim_{n \to \infty} 0 = 0,$$
in other words $x = 0$ for all $x \in A^{\perp}$.
Best Answer
This is not a Hilbert space. Define $$ f_n(x) = \max(0, \min(nx,1)). $$ Then $f_n(x) \to 1$ for $x>0$ and $f_n(x)=0$ for all $x<0$. However $(f_n)$ is a Cauchy sequence in the given norm. Let $n>m\ge1$. Then $$ \|f_n-f_m\|^2 = \sum_{k: \ r_k \in (0,\frac1m)} 2^{-k} |f_n(r_k)-f_m(r_k)|^2 \le \sum_{k: \ r_k \in (0,\frac1m)} 2^{-k}. $$ It remains to show that the right-hand side converges to zero for $m\to\infty$. Take $g$ with $g(x)=1$ for all $x$. Then $$ 1 = \|g\|^2 \ge \sum_{k: \ r_k \in (0,1)} 2^{-k} = \sum_{i=1}^\infty \sum_{k: \ r_k \in [\frac1{i+1},\frac1i)} 2^{-k}. $$ The sum $\sum_{k: \ r_k \in (0,\frac1m)} 2^{-k}$ is a tail of this converging series, hence $\sum_{k: \ r_k \in (0,\frac1m)} 2^{-k} \to 0$ for $m\to\infty$. Thus, $(f_n)$ is Cauchy. The sequence converges pointwise on the rationals to a function that cannot be extended to be a continuous function.