Answer :
Let [tex]\eta\in H\cap K[/tex]. Since both [tex]H[/tex] and [tex]K[/tex] are subgroups of [tex]G[/tex], we have
[tex]\eta\in H\cap K\implies\begin{cases}\eta\in H\\\eta\in K\end{cases}\implies\eta\in G[/tex]
Because both [tex]H,K[/tex] are subgroups of [tex]G[/tex], then [tex](H\cap K,\bullet)[/tex] contains the identity element [tex]e[/tex]. Furthermore, there must be some element [tex]\eta^{-1}\in H[/tex] and [tex]\eta^{-1}\in K[/tex], i.e. [tex]\eta^{-1}\in H\cap K[/tex], such that [tex]\eta\bullet\eta^{-1}=e[/tex].
Now let [tex]\eta_1,\eta_2,\eta_3\in H\cap K[/tex]. By the same reasoning as above it follows that each of these belong in [tex]G[/tex], and since [tex](G,\bullet)[/tex] is a group, we have [tex]\eta_1\bullet(\eta_2\bullet\eta_3)=(\eta_1\bullet\eta_2)\bullet\eta_3[/tex], so [tex]H\cap K[/tex] is associative under [tex]\bullet[/tex].
So [tex]H\cap K[/tex] contains the identity, is closed with respect to inverses, and is associative under [tex]\bullet[/tex]. Therefore [tex]H\cap K[/tex] must be a subgroup of [tex]G[/tex].
In the case of union, this is not always the case. Consider the group [tex](\mathbb Z,+)[/tex] with subgroups [tex](H,+)[/tex] and [tex](K,+)[/tex] where [tex]H[/tex] is the set of all integer multiples of 2 and [tex]K[/tex] is the set of all integer multiples of 3.
It's easy to show that [tex]H[/tex] and [tex]K[/tex] are indeed subgroups, but this is not the case for [tex]H\cup K[/tex]. We have
[tex]H\cup K=\{0,\pm2,\pm4,\ldots\}\cup\{0,\pm3,\pm6,\ldots\}[/tex]
Take the elements [tex]2\in H[/tex] and [tex]3\in K[/tex]. Addition yields [tex]2+3=5[/tex], but [tex]5\not\in H\cup K[/tex], so [tex]H\cup K[/tex] is not closed under [tex]+[/tex].
[tex]\eta\in H\cap K\implies\begin{cases}\eta\in H\\\eta\in K\end{cases}\implies\eta\in G[/tex]
Because both [tex]H,K[/tex] are subgroups of [tex]G[/tex], then [tex](H\cap K,\bullet)[/tex] contains the identity element [tex]e[/tex]. Furthermore, there must be some element [tex]\eta^{-1}\in H[/tex] and [tex]\eta^{-1}\in K[/tex], i.e. [tex]\eta^{-1}\in H\cap K[/tex], such that [tex]\eta\bullet\eta^{-1}=e[/tex].
Now let [tex]\eta_1,\eta_2,\eta_3\in H\cap K[/tex]. By the same reasoning as above it follows that each of these belong in [tex]G[/tex], and since [tex](G,\bullet)[/tex] is a group, we have [tex]\eta_1\bullet(\eta_2\bullet\eta_3)=(\eta_1\bullet\eta_2)\bullet\eta_3[/tex], so [tex]H\cap K[/tex] is associative under [tex]\bullet[/tex].
So [tex]H\cap K[/tex] contains the identity, is closed with respect to inverses, and is associative under [tex]\bullet[/tex]. Therefore [tex]H\cap K[/tex] must be a subgroup of [tex]G[/tex].
In the case of union, this is not always the case. Consider the group [tex](\mathbb Z,+)[/tex] with subgroups [tex](H,+)[/tex] and [tex](K,+)[/tex] where [tex]H[/tex] is the set of all integer multiples of 2 and [tex]K[/tex] is the set of all integer multiples of 3.
It's easy to show that [tex]H[/tex] and [tex]K[/tex] are indeed subgroups, but this is not the case for [tex]H\cup K[/tex]. We have
[tex]H\cup K=\{0,\pm2,\pm4,\ldots\}\cup\{0,\pm3,\pm6,\ldots\}[/tex]
Take the elements [tex]2\in H[/tex] and [tex]3\in K[/tex]. Addition yields [tex]2+3=5[/tex], but [tex]5\not\in H\cup K[/tex], so [tex]H\cup K[/tex] is not closed under [tex]+[/tex].