I need to prove that $M = S^3 \setminus S^1$ doesn't admit any metric of constantly negative sectional curvature s.t. $M$ is complete respect for this metric. I know that it is consequence of famous Thurston Theorem, but it is quite uncomfortable to use it, so, maybe there is exist direct argument?

3$\begingroup$ You need to assume that the $S^1$ is unknotted. Then consider the limit set of the underlying Kleinian group of $S^3S^1$. It will contain at most 2 points, and so couldn't be a finite covolume Kleinian group. $\endgroup$ – Autumn Kent Jun 6 '18 at 22:45

2$\begingroup$ I am not sure why people offer complicated answers. Your $M$ does admits a complete hyperbolic metric, namely, the quotient of the upper half space $\mathbf{H^3}=\{(z,t): z\in\mathbb C, t>0\}$ by the cyclic group generated by the dilation $(z,t)\to (2z, 2t)$. $\endgroup$ – Igor Belegradek Jun 7 '18 at 20:07
You probably mean $M$ does not admit complete hyperbolic metrics of finite volume.
Since $M$ is topologically the interior of a solid torus, a complete hyperbolic structure just identifies $M$ as the quotient of $\mathbb{H}^3$ by a single loxodromic or parabolic isometry, but such quotients have infinite volume.

$\begingroup$ Thank you! Does exist such simple explanation in case of infinite volume too? $\endgroup$ – kp9r4d Jun 5 '18 at 5:24
The following contribution comes from conversations with Bill Goldman, any mistakes however are mine alone.
Any (geodesically complete) geometric 3manifold $N=\mathbb{M}/G$ with infinite order elements in its fundamental group (isomorphic to $G$) is covered by an open solid torus admitting the same geometric structure (having model geometry $\mathbb{M}$). Note that $G$ is a subgroup of the isometry group of $\mathbb{M}$ and arises from a discrete faithful homomorphism $\pi_1(N)\to Isom(\mathbb{M})$.
To see this, let $T\in G$ be of infinite order. Then the covering space of $N$ given by $\mathbb{M}/\langle T\rangle$ is an open solid torus (coverings of manifolds admitting geometric structures admit geometric structures themselves). Although I have not verified it, I am pretty sure that $\mathbb{M}$ can be any of Thurston's eight geometries excepting only $\mathbb{S}^3$.
Here is an example construction (when $\mathbb{M}$ is hyperbolic 3space $\mathbb{H}^3$). Take a hyperbolic line $L$ (complete geodesic) in $\mathbb{H}^3$ and a hyperbolic translation (loxodromic element) along that line. Call the translation $T$. Consider the cyclic group $\langle T\rangle.$ Then $T$ leaves the line $L$ invariant.
Take any unit disk $D$ orthogonal to $L$. Translate $D$ by $T$ along $L$ to obtain the disk $T(D)$. The two disks $D$ and $T(D)$ bound a fundamental domain for $\langle T\rangle$.
Fundamental Domain in Ball Model; Image made by Marvin Castellon
The quotient of the fundamental domain by $\langle T\rangle$ is homeomorphic to a solid torus (one can do the same thing when $T$ is parabolic). The resulting manifold is geodesically complete.
The volume of such a torus must be infinite. For otherwise, by Mostow Rigidity, there would be a unique hyperbolic structure. However, from the construction above we see there are infinitely many distinct structures (pick any nonconjugate loxodromic element).
So in your case the manifold is topologically an open solid torus (assuming $S^1$ is unknotted) and so it admits many different complete geometric structures (including many hyperbolic structures).