Closure properties

I is false: for instance, consider $S=[0,1)$ and $T = (1,2]$ inside $X={\mathbb R}.$ The intersection of the closures is $[0,1] \cap [1,2] = \{1\},$ but the closure of the intersection is the closure of the empty set, which is the empty set.

III is false: for instance, the closure of $\mathbb R\setminus \{0\}$ is $\mathbb R.$ (There are even more dramatic examples; for instance, the closure of $\mathbb Q$ is $\mathbb R$ as well.)

II is true. There are several ways to prove it; one way is to look at complements. This statement is equivalent to the statement that the interior of $U \cap V$ equals the intersection of the interior of $U$ and the interior of $V.$ Here is a brief proof of this equivalent statement:

If $x$ is in the interior of $U \cap V,$ there is an open ball around it that is contained entirely in $U\cap V,$ so it is contained in $U$ and in $V,$ so $x$ is in the interior of $U$ and the interior of $V.$ On the other hand, if $x$ is in the interior of $U$ and the interior of $V,$ there are two open balls around $x,$ one of which is contained in $U$ and the other in $V.$ The intersection of these two open balls is contained in $U \cap V,$ so $x$ is in the interior of $U \cap V.$ (Note that this argument only works with finitely many sets; II becomes false if we replace the union of two sets with the union of infinitely many.)

I'm sure there are more elegant ways to prove II directly, since there are lots of different equivalent definitions of closure.

$\boxed {1.-}$ Let $X = \mathbb{R}$ with the usual distance and $S = \{\text{ rational numbers }\}$ and $T = \{\text{irrational numbers}\}$ then $\varnothing = S \cap T = \overline{ S \cap T} \neq \overline{S} \cap \overline{T} = \mathbb{R} \cap \mathbb{R} = \mathbb{R} \Rightarrow$ I is false.

$\boxed {2.-}$ $\begin{cases} S \subseteq S \cup T \Rightarrow \overline{S} \subseteq \overline{S \cup T} \\ T \subseteq S \cup T \Rightarrow \overline{T} \subseteq \overline{S \cup T} \end {cases} \Rightarrow \overline{S} \cup \overline{T} \subseteq \overline {S \cup T}.$ If $x \in \overline{S \cup T}$ then $\forall \epsilon > 0$ and for all ball $B(x, \epsilon)$ is satisfied $B(x, \epsilon) \cap ( S \cup T) \neq \varnothing \Rightarrow$ $B(x, \epsilon) \cap S \neq \varnothing$ or $B(x, \epsilon) \cap T \neq \varnothing$ and this implies that $x \in \overline{S}$ or $x \in \overline{T}$ , ( $\overline{S \cup T} \subseteq \overline{S} \cup \overline{T}$ ) (use reductio ad absurdum) and hence II is true.

$\boxed{3.-}$ $S = \{\text{rational numbers}\} \Rightarrow \overline{S} = \mathbb{R}$ and $S \neq \mathbb{R}$ and therefore III is false.