To finish your argument for (1) implies (2), we pick $x_n \in X\setminus \cup_{i=1}^n O_i$, then let $(y_n)$ be a subsequence of $(x_n)$ which converges to $y\in X$. Since the $O_i$ are an open cover of $X$, there is some $N$ so that $y\in O_N$. But there are infinitely many $y_n \notin O_N$, which is impossible: if $(y_n)\to y$, then for any open neighborhood $U$ of $y$, there must be some $m$ so that for $n\ge m$, we have $y_n\in U$.
To prove that (2) implies (1) without completeness, let me define a limit point of $Y=\{f_n\}$: a point $z\in H$ is a limit point of $Y$ if for every open neighborhood $U$ of $z$, we have that $(U\setminus \{z\}) \cap Y \ne \emptyset$. It's easy to show that in any topological space, the closure of a set $Y$ is equal to $Y$ along with the limit points of $Y$. Moreover, if $z$ is a limit point of $Y = \{f_n\}$, then it's easy to show that there is a subsequence of $(f_n)$ which converges to $z$: pick $n_1\in\mathbb{N}$ so that $f_{n_1} \in (B_{1}(z)\setminus \{z\}) \cap Y$, then let $\delta_1 = \mbox{dist}(z,f_{n_1})$ and pick $n_2\in \mathbb{N}$ so that $f_{n_2} \in (B_{\delta_1/2}(z)\setminus \{z\}) \cap Y$, and so on. This subsequence $(f_{n_k})$ converges to $z$. So now, as you say, this closure, which you called $X'$, is compact. If $X'$ contains any limit points of $Y$, then we're done. Otherwise, $X' = Y$ and there are no limit points of $Y$. Thus, for each $n\in\mathbb{N}$, there is an open neighborhood $U_n$ of $f_n$ with $Y\cap U_n = \{f_n\}$. Now the $U_n$ form an open cover of $Y$ with no finite subcover.
You're right! We didn't use much about the Hilbert space $H$. It's true for all metric spaces. Though you should change your second definition of compactness to
For any sequence $(x_n)$ in $X$, there is some a subsequence $(y_n)$ and some $y\in X$ such that for any open neighborhood $U$ of $y$, there is some $m\in \mathbb{N}$ so that if $n\ge m$, then $y_n\in U$.
It's not true for all topological spaces. Not even for Hausdorff topological spaces.
Best Answer
The comment by Eric Wofsey is correct: If $C$ is compact, then $C(A)$ being precompact means that $\overline{C(A)}$ is compact.
Let $C: H \to H$ be a linear operator on $H$, and suppose $C$ is a compact operator. Now, let $A = \{x_n\} \subset H$ be a bounded sequence. Since $A$ is bounded, the image sequence $C(A) = \{Cx_n\}$ is precompact. Hence, $\overline{C(A)}$ is compact. At this point, if you know compactness in a metric space is equivalent to every sequence having a convergent subsequence, we are done because $\{Cx_n\}$ is a sequence in the compact space $\overline{C(A)}$.