I've looked at a number of references on this including some questions on stack exchange. Am I correct if I summarize by stating the following ?
(1) A space C (metric, normed, or inner product) is complete iff it contains the limit points of all Cauchy sequences in the space.
(2) A subspace S of V is closed iff it contains the limit points of all Cauchy sequences in the subspace S which exist in V itself. So, if V is not complete then S can be closed even though it doesn't contain the limit points of all its Cauchy sequences (provided they don't exist in V).
I can see if that is correct, that a closed subspace of a complete space is complete. Also, that an incomplete subspace can be closed: for example, the interval [0, 1] is closed in Q (the rationals) without being complete.
Confirmation or correction would be much appreciated.
(1) is the definition of "complete". (2) is a consequence of the definitions, but is not how "closed" is defined.
Completeness is only definable for metric spaces (or certain other spaces with a similar structure). Normed and inner product spaces are just examples of metric spaces where the metric is defined in terms of other structure on the set.
Closed, however, as a much broader definition. It applies to all topological spaces, including ones that have no metric and behave in ways that defy all intuition. In general, we define
$$\text{A subset } C \text{ of a topological space }X\text{ is closed if and only if its complement }X \setminus C\text{ is open.}\tag{1}$$
For metric spaces, a set $A$ is open if every point $x$ of the set has some ball $B_\epsilon(x) = \{ y \in A\ |\ d(x, y) < \epsilon\}$ around it which is also contained in the set. This means that, since the complement of a closed set is an open set, $$\text{A subset }C\text{ of a metric space }M\text{ is closed if and only if for every }x \notin C\text{, there is some ball }B_\epsilon(x)\text{ such that }B_\epsilon(x) \cap C = \emptyset\tag{2}.$$ That is, not only is $x$ not in $C$, but once you look near enough to $x$, there is nothing from $C$ nearby. Now obviously such an $x$ cannot be the limit of a sequence in $C$, since if $\lim_n x_n = x$, then by definition of the limit, every neigborhood of $x$ must contain some elements of $\{x_n\}$. Hence the limit of any convergent sequence in a closed set $C$ must itself be contained in $C$.
Conversely, suppose $D$ is a subset of $M$ such that every convergent sequence in $D$ has its limit in $D$. Let $x \in M \setminus D$. Now one or the other of the following must be true:
If the first statement is true, then for that value of $\epsilon, B_\epsilon(x) \cap D = \emptyset$, as $x \notin D$. Now examine the case where the second statement is true. Suppose that for all $\epsilon>0, B_\epsilon(x) \cap D \ne \emptyset$. Then for each $n \in \Bbb N$ we can choose $x_n \in B_{1/(1+n)}(x) \cap D$. Then $\lim_n x_n = x$, but $\{x_n\} \subseteq D$, and so $x \in D$. A contradiction. Therefore the supposition that for all $\epsilon>0, B_\epsilon(x) \cap D \ne \emptyset$ must be false. Thus, no matter which of the two cases listed above holds, there is a ball $B_\epsilon(x)$ such that $B_\epsilon(x) \cap D = \emptyset$. Thus by closure condition (2), $D$ is closed. This gives: $$\text{A subset }C\text{ of a metric space }M\text{ is closed if and only if every convergent}$$ $$\text{sequence in }C\text{ has its limit in }C.\tag{3}$$
Finally, any convergent sequence in a metric space is Cauchy, and a Cauchy sequence that has a limit point necessarily converges to that limit point. This provides your formulation.