Instead of saying "$A$ is sequentially closed", the conclusion of the problem really should say that $A$ is "weakly sequentially closed" or "sequentially weakly closed."
The problem is asking to show that closed in the weak topology implies sequentially closed in the weak topology.
A solution can be given that does not use the specifics of the weak topology at all. For any topological space, closed implies sequentially closed.
Let $X$ be any topological space and let be $A$ any closed subset of $X$. We need to show that $A$ is sequentially closed. Let $x \in X$ and $(x_n) \subseteq A$ be such that $x_n \to x$. We need to show $x \in A$. Let $U$ be an arbitrary open neighbourhood of $x$. Since $x_n \to x$, we have $x_n \in U$ for all but finitely many $n$, and so $U$ contains at least one point of $A$. Since $U$ was an arbitrary open neighbourhood of $x$, this proves $x$ is in the closure of $A$. Since $A$ is closed, $x$ is in $A$.
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.
Best Answer
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.