Let $G$ be a locally compact group and $L^2G$ denote the Hilbert space of complex functions that are square measurable with respect to the Haar measure.
Let $\lambda : G \to U(L^2G)$ denote the left regular representation. It is well-known that $\lambda$ is faithful.
The question: is $\lambda$ an open map? If not, is it open at least when $G$ is compact?
Definitely not. The easiest way to see this is to realize that for a measure space $(X,\mu)$ any measure preserving map $\phi:X\rightarrow X$ gives rise to an element of $U(L^2(X,\mu))$, the case of the left regular representation is just left translation on the group, which preserves the Haar measure. So to see that the image of $G$ in $U(L^2G)$ is not open just take two sequences of disjoint measureable sets in $G$ with $\cdots\subset A_2\subset A_1\subset A_0 = A$ and $\cdots\subset B_2\subset B_1\subset B_0 = B$ such that $A\cap B=\emptyset$ and $\lim \mu(A_n)=\lim \mu(B_n) = 0$ and $\mu(A_i) = \mu(B_i)$ for all $i$. Then let $\phi_n$ be measurable maps which swap $A_n$ and $B_n$ but are the identity everywhere else.
Then the corresponding unitary maps will converge to the identity but you can choose this so that the $\phi_n$ are not homomorphisms of the group and so lie in the complement of $G$ in $U(L^2G)$. Thus the complement is not closed so the image is not open. Compactness of $G$ is not relevant.
More generally, this idea shows that the image of $G$ is actually very far from being open, in fact it is nowhere dense in general.
One last fact to point out. Note that as long as $G$ is $\sigma$-finite with respect to the Haar measure then $L^2$ is the unique separable Hilbert space (if $G$ is infinite). So $U(L^2G)$ doesn't depend on $G$! Also every other second countable locally compact groups sits inside $U(L^2G)$.