All closed hyperbolic 3-manifold groups embed into a compact Lie group. 

To prove this, note first of all that given a hyperbolic 3-manifold $M$, it suffices to show that a finite-index subgroup $G\leq \pi_1(M)$ of index $m$ embeds into a compact Lie group. Then the representation $\rho: G\hookrightarrow O(n)$ will [induce a represenation][1] $Ind_G^{\pi_1(M)} \rho : \pi_1(M) \hookrightarrow O(nm)$. 

Now, by the [proof of the virtual Haken conjecture][2], $\pi_1(M)$ has a finite-index subgroup $G$ which is the fundamental group of a [special cube complex][3], which implies that $G$ embeds into a [right-angled artin group][4] $A$, [and hence into a right-angled Coxeter group][5] $\Gamma$. 

Finally, right-angled Coxeter groups have faithful embeddings into $O(n)$. This follows from a [result of Vinberg][6], which gives a faithful linear action by reflections on $\mathbb{R}^n$. We'll review this construction following section 7 of [this paper.][7] 


Fix a right-angled Coxeter group
$$\Gamma = \langle \gamma_1,\dots,\gamma_k ~|~ (\gamma_i \gamma_j)^{m_{i,j}} = 1\quad \forall i,j\rangle,$$
where $m_{i,i}=1$ and $m_{i,j}\in\{ 2,\infty\}$ for all $i\neq j$.


For $t \in \mathbb{R}$, the matrix $M_t=(M_t(i,j))_{1\leq i,j\leq k}$ where
$$M_t(i,j) = \left\{ \begin{array}{cl}
1 & \text{if }m_{i,j}=1, \text{ i.e.  $i=j$},\\
0 & \text{if }m_{i,j}=2,\\
-t & \text{if }m_{i,j}=\infty
\end{array}\right.$$
 defines a symmetric bilinear form $\langle\cdot,\cdot\rangle_t$ on $\mathbb{R}^k$.
Note that $\mathrm{det}(M_t)$ is a nonzero polynomial in $t$ (take $t=0$), hence it is nonzero outside of some finite set $F$ of exceptional values of $t$.
For any $t\in \mathbb{R}-F$, the form $\langle\cdot,\cdot\rangle_t$ is nondegenerate.
Define the representation $\rho_t : \Gamma\to\mathrm{Aut}(\langle\cdot,\cdot\rangle_t)\leq \rm{GL}(k,\mathbb{R})$ by
$$\rho_t(\gamma_i) : v \mapsto v - 2\langle v, e_i \rangle_t \, e_i $$
for all $i$. Each generator is a reflection in a hyperplane perpendicular to $e_i$ with respect to the metric $\langle\cdot,\cdot\rangle_t$.

For $t>1$, the convex cone
$$\widetilde{\Delta}_t = \{ v\in\mathbb{R}^k ~|~ \langle v, e_i\rangle_t \leq 0 \ \,\forall i\}$$
descends to a convex polytope $\Delta_t$ in an affine chart of $\mathbb{P}(\mathbb{R}^{k})$.
By Theorem 2 of [Vinberg][6], the representation $\rho_t$ is discrete, faithful, the set $\rho_t(\Gamma)\cdot\Delta_t$ is convex in $\mathbb{P}(\mathbb{R}^k)$, and the action of $\Gamma$ on the open set
$$\mathcal{U}_t:=\mathrm{Int} \left ( \rho_t(\Gamma)\cdot\Delta_t \right )$$ is properly discontinuous.

Now, let $t$ be close to $0$ and transcendental, so that $M_t$ is positive definite. Then $\rho_t : \Gamma\to \mathrm{Aut}(\langle\cdot,\cdot\rangle_t) \cong O(k)$ is faithful (it is Galois conjugate to a representation for some transcendental $t>1$). 

**Remark:** I think the first step of taking an induced representation can be eliminated. If we assume $G\lhd \pi_1(M)$, and $K=\pi_1(M)/G$, then one may embed $G$ into a right-angled Coxeter group which admits an action of $K$ by permuting its generators, and so that the embedding of $G$ is $K$-equivariant. Then $\pi_1(M)$ should embed in this Coxeter group extended by $K$, which clearly still also has a faithful representation.

  [1]: https://en.wikipedia.org/wiki/Induced_representation
  [2]: https://www.math.uni-bielefeld.de/documenta/vol-18/33.html
  [3]: https://link.springer.com/article/10.1007%2Fs00039-007-0629-4
  [4]: https://en.wikipedia.org/wiki/Artin_group#Right-angled_Artin_groups
  [5]: https://mathscinet.ams.org/mathscinet-getitem?mr=1783167
  [6]: https://www.turpion.org/php/paper.phtml?journal_id=im&paper_id=1203
  [7]: https://arxiv.org/abs/1804.03132