Since $x$ is continuous the image of $[0,N]$ is is norm bounded for each positive integer $N$. Since $\lim_{t \to \infty} \|x(t)-y_0\|$ exists it follows that $\{x(t): t \geq 0\}$ lies in a norm bounded subset of $H$. In a separable Hilbert space any norm bounded set is weakly sequentially precompact.
The thesis is:
$$\int_\Omega (n_k\nabla V_k-n\nabla V)\cdot\nabla\varphi\;\text{d}x \to 0$$
Observe first:
$$\int_\Omega (n_k\nabla V_k-n\nabla V)\cdot\nabla\varphi\;\text{d}x =
\int_\Omega (n_k-n)\nabla V_k\cdot\nabla\varphi\;\text{d}x
+ \int_\Omega (\nabla V_k-\nabla V)\cdot n \nabla\varphi\;\text{d}x
$$
The second term goes to zero thanks to $V_k\to V$ weakly in $H^1$. This is because the map:
$$\psi \mapsto \int_\Omega \nabla \psi\cdot n \nabla\varphi\;\text{d}x$$
is linear and continuous on $H^1$.
For the first term, the argument is a bit longer and uses two key ideas: the compact embedding of $H^1$ in $L^2$ and the density of $C^{\infty}(\mathbb{R}^d)$ in $H^1$.
$H^1$ embeds compactly in $L^2$. For this reason $n_k\to n$ weakly in $H^1$ entails $n_k\to n$ strongly in $L^2$. This can be used to show:
$$\int_\Omega (n_k-n)\nabla V_k\cdot\nabla\phi\;\text{d}x \to 0$$
for every $\phi \in C^{\infty}(\mathbb{R}^d)$.
In what follows it is explained in details how to use the density argument. Observe that:
$$\bigg| \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\psi\;\text{d}x \bigg| \leq 2 A B \|\psi\|_{H^1} \qquad \forall \psi \in H^1$$
where $A,B$ are positive constants satisfying $\|n_k\|_\infty , \|n\|_\infty \leq A$ and $\|V_k\|_{H^1} \leq B$ for every $k$. Such constants existence is granted by boundedness of $n_k$ and $V_k$ in $L^{\infty}$ and $H^1$ respectively. Therefore,
for every $\phi \in C^{\infty}(\mathbb{R}^d)$ we have:
$$\bigg| \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\varphi\;\text{d}x \bigg| \leq \bigg| \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\phi\;\text{d}x \bigg| + 2 A B \|\varphi - \phi\|_{H^1}$$
which yields:
$$ \limsup_{k\to \infty}\bigg| \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\varphi\;\text{d}x \bigg| \leq 2 A B \|\varphi - \phi\|_{H^1}$$
Since $\|\varphi - \phi\|_{H^1}$ can be chosen arbitrarily small thanks to $C^{\infty}(\mathbb{R}^d)$ density in $H^1$, this concludes the proof.
Best Answer
Since $x : [0,\infty) \to H$ is continuous, and it was shown that $\lim_{t \to \infty} \|x(t)-y_0\|$ exists, then in particular, $\{x(t) : t \ge 0\}$ is bounded in norm. So by Alaoglu's theorem, it is weakly compact.
This also implies weak sequential compactness. Indeed, if $H$ is separable then any bounded set is weakly metrizable, so then weak compactness and weak sequential compactness are equivalent. If $H$ is not separable, then given a sequence $\{x(t_n)\}$, consider the separable subspace $H' \subset H$ consisting of the closed linear span of $\{x(t_n)\}$. Then by the previous case there is a subsequence converging weakly in $H'$, which thus also converges weakly in $H$.
After that, I think you're misreading the sentence around (1.5). What is being asserted in this sentence (and what the rest of the proof goes on to show) is the conditional statement that if $x(t_n)$ and $x(s_n)$ both converge weakly then their limits must be equal.
To see why this assertion (call it A) implies the claim that $x(t_n)$ converges weakly, use the "double subsequence trick". Suppose it did not. Weak compactness says there is a subsequence $t_{n_k}$ (which is also a (*)-sequence) with $x(t_{n_k})$ converging weakly to some $x^*$. Since the original sequence $x(t_n)$ did not converge weakly to $x^*$ or anything else, there is a weakly open neighborhood $U$ of $x^*$ and a subsequence $t_{m_k}$ such that all $x(t_{m_k})$ are outside $U$. This subsequence has a further subsequence converging weakly to some $x^{**}$, which is necessarily outside $U$ and in particular not equal to $x^*$. But this contradicts the assertion A.