Is it right that $\cup_{g\in G}gHg^{-1}\subsetneq G$, for every $H$, proper subgroup of infinite group $G$? Let $G$ be a group and $H$, a proper subgroup of $G$ with $[G:H]<\infty$. Is it  right that $\bigcup\limits_{g\in G}gHg^{-1}$ is a proper subset of $G $? If $G$ is a finite group we can prove this claim, but is this right for infinite groups?

Yes it is correct. Consider the action of $G$ by left multiplication on the finite set $X$ of left cosets of $H$ in $G$. This action is transitive. The action defines a homomorphism $\phi:G \to {\rm Sym}(X)$.
Then image $P$ of $\phi$ is a transitive group group on the finite set $X$, and there must exist an element $a \in P$  without fixed points. (This follows from the result that the average number of fixed points in a transitive action on a finite set is $1$, and the identity fixes more than one point - since $H$ is a proper subgroup we have $|X| > 1$).
Now, if $b \in gHg^{-1}$ for some $g \in G$, then $b(gH) = gH$, so $\phi(b)$ fixes $gH$. Hence an element $b \in G$ with $\phi(b) = a$ does not lie in $\cup_{g \in G} gHg^{-1}$.
