= Solution
The convention in this question counts a point whenever every open <neighbourhood> meets $A$, without excluding the point itself. These are <adherent points>, whose set is the <closure> $\overline A$; they differ from <limit points> under the convention that requires a different point of $A$ in each neighbourhood.
If $A$ is a <closed set> and $x\notin A$, the <open set> $X\setminus A$ is a neighbourhood of $x$ disjoint from $A$. Thus every adherent point lies in $A$. Conversely, if every adherent point lies in $A$, each $x\notin A$ has an open neighbourhood disjoint from $A$. Their union is $X\setminus A$, which is open. Hence \b[$A$ is closed exactly when it contains all the points specified in the question].
The <interior of a set> and <closure> are
$$
\boxed{\operatorname{Int}(A)=\bigcup\{U:U\text{ open},\ U\subseteq A\},\qquad
\overline A=\bigcap\{F:F\text{ closed},\ A\subseteq F\}.}
$$
The neighbourhood description of the <closure> follows because failure to be adherent is equivalent to lying in an open set disjoint from $A$.
For <closure of a connected set>, suppose $\overline A=U\sqcup V$ were a separation into two nonempty relatively open sets. Their intersections with the <connected subset> $A$ would separate $A$ unless one intersection were empty. Say $A\subseteq U$. Choose $v\in V$ and an open set $O\subseteq X$ with $v\in O$ and $O\cap\overline A\subseteq V$. Since $v\in\overline A$, the set $O$ meets $A$, contradicting $A\subseteq U$. Thus \b[the closure of a connected set is connected]. The empty-set case holds directly.
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