= Solution
The <intersection> $H\cap K$ contains the <identity element>. If $a,b\in H\cap K$, then $ab^{-1}$ belongs to both $H$ and $K$, so it belongs to their intersection. The <subgroup test> proves that $H\cap K$ is a <subgroup> of $G$, with the inherited operation. Explicitly, closure and inverses come from this test, and <associativity> is inherited from $G$.
Label the vertices of the <equilateral triangle> $1,2,3$ counterclockwise. For the <dihedral group> of order six, let $r=(123)$ be the counterclockwise rotation by $120^\circ$, and let $s=(23)$ be reflection in the line through vertex 1 and the opposite side's midpoint. Choose
$$
I=e,\quad A=r,\quad B=r^2,\quad C=s,\quad D=rs,\quad E=r^2s.
$$
Then $B$ rotates by $240^\circ$. The three <Euclidean reflections> $C,D,E$ fix vertices $1,3,2$ respectively. Products act on a vertex from right to left. The relations $r^3=s^2=e$ and $sr=r^{-1}s$ give
$$
(r^is^\epsilon)(r^js^\eta)=r^{\,i+(-1)^\epsilon j}s^{\epsilon+\eta},
$$
where exponents of $r$ are reduced modulo three and those of $s$ modulo two. This constructs the entire <Cayley table>:
$$
\begin{array}{c|cccccc}
&I&A&B&C&D&E\\\hline
I&I&A&B&C&D&E\\
A&A&B&I&D&E&C\\
B&B&I&A&E&C&D\\
C&C&E&D&I&B&A\\
D&D&C&E&A&I&B\\
E&E&D&C&B&A&I
\end{array}.
$$
For example $AC=D$ whereas $CA=E$, so the group is \b[nonabelian].
A subgroup of order two is generated by its unique nonidentity element, which must have order two. The rotations $A,B$ have order three, and each reflection has order two. Thus there are exactly
$$
\boxed{\langle C\rangle=\{I,C\},\qquad\langle D\rangle=\{I,D\},\qquad\langle E\rangle=\{I,E\}.}
$$
None is a <normal subgroup>. Conjugation by $A$ sends the reflection $r^js$ to $r^{j+2}s$, cycling the three reflections. For example $ACA^{-1}=E\notin\langle C\rangle$. Each of the other two subgroups fails the same conjugation test.
Finally, take two distinct such subgroups. For instance their <union> is $\langle C\rangle\cup\langle D\rangle=\{I,C,D\}$, but $CD=B$ lies outside it. Thus it is \b[not closed under the inherited group operation] and is not a <group>. The distinctness is necessary: the union of a subgroup with itself is that subgroup.
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