Here is my derivation:
An element in $SO(3)$ can be written as $(\vec{v}_1,\vec{v}_2,\vec{v}_3)^T$ with $\{\vec{v}_i\}$ an oriented orthonormal set of basis. The set of possible $\vec{v}_1$ forms an $S^2$, and the choice of $\vec{v}_2,\vec{v}_3$ forms an $S^1$. Therefore, $SO(3)$ is just $S^2\times S^1$.
I know this must be incorrect since $\pi_1(SO(3))=\mathbb{Z}_2$ while $\pi_1(S^2\times S^1)=\mathbb{Z}$, but I don't know where did I go wrong.
I don't think any of the other answers really address why your argument doesn't prove that $SO(3)$ is just $S^2 \times S^1$.
Your argument seems reasonable, but there's a crucial flaw. The idea is that we pick a point $p \in S^2$ to identify with $\vec{v_1}$ Fine so far. Now we pick an angle $\theta \in S^1$ which sets the direction of $\vec{v_2}$. It seems like we should be able to do this, but we can't because of the topological properties of $S^2$.
More explicitly, let $\phi : S^2 \times S^1 \to SO(3)$ be your proposed isomorphism, where, as above, $(p, \theta) \mapsto (\text{the ONB with $\vec{v_1}$ at $p$ and with $\vec{v_2}$ at angle $\theta$})$. Consider the map $f : S^2 \to SO(3)$ given by $f(p) = \phi(p, 0)$. That is, $f(p)$ is the ONB with $\vec{v_1^0}(p)$ pointing to $p$ and $\vec{v_2^0}(p)$ at angle $0$. Believe it or not, we've already gone wrong. We've made a continuous choice of tangent vector $\vec{v_2^0}(p)$ to $S^2$ at every point $p$ of $S^2$. In other words, we've attached a non-vanishing tangent vector field to every point of a sphere. The famous hairy ball theorem says that this is impossible. What to conclude? Our original isomorphism must not have actually been well-defined.
What went wrong? It was where we said that $\vec{v_2}$ should be at angle $\theta$. With respect to what? Making a choice of direction with reference to which we measure $\theta$ at each point of $S^2$ is precisely attaching a non-vanishing tangent vector field to every point of the sphere (the vector field that points in the direction $\theta = 0$ at every point). But the hairy ball theorem says this is impossible.
(What follows will be a sketch. Consult a textbook on fiber bundles for detail and rigor)
Can we fix up your argument? Yes! This is one of the motivating examples for a mathematical structure called a "fiber bundle." We would like to attach $S^1$ to every point of $S^2$, but in a way where it actually does describe $SO(3)$. The resulting space will not be $S^2 \times S^1$, but instead something we call an $S^1$ (or circle) bundle over $S^2$.
We'll have two measurements of angle, $\theta_1$ and $\theta_2$. $\theta_1$ measures the angle between $\vec{v_2}$ and north. This will be a good $S^1$ coordinate everywhere except the north and south pole where there is no well-defined north direction. To make up for this, we'll have to have a $\theta_2$ which we'll arbitrarily decide is the angle between $\vec{v_2}$ and the direction you'd go to get to the intersection of the equator and prime-meridian as quickly as possible (here we are imagining $S^2$ as the Earth).
We have now attached to all points of $S^2$ a copy of $S^1$ with a well-defined $0$ direction. Unfortunately, we've attached two copies of $S^1$ to all but $4$ points, and they don't agree on which direction $0$ is: one copy is "twisted" with respect to the other. But this is not a serious problem. It's very easy to convert between one way of measuring angle and the other, and in fact this can be done smoothly at each point that the two copies of $S^1$ overlap.
By attaching $S^1$ in this "twisted" way, we've turned $S^2$ into a non-trivial fiber bundle. This fiber bundle really is isomorphic (in the right category) to $SO(3)$, and the proof is precisely the argument you sketched in your post. An element of $SO(3)$ is an ONB with $\vec{v_1}$ pointing to a point of $S^2$ and $\vec{v_2}$ at angle $\theta_1$ and/or $\theta_2$ measured as described above.
Consider the quaternions $\Bbb H = \mathrm{span}\{1,i,j,k\} \simeq \Bbb R^4$, which is nothing else but $\Bbb R^4$ endowed with the algebra structure given by $i^2=j^2=k^2=ijk=-1$. For $q= a+ib+jc+kd\in \Bbb H$, we define its conjugate $\bar{q}=a-ib-jc-kd$ and its real part $\mathrm{Re}(q) = a = \frac{1}{2}(q+\bar{q})$. The usual inner product on $\Bbb R^4$ reads $\langle q_1,q_2\rangle = \mathrm{Re}(q_1\bar{q_2})$, and the associated norm is $\|q\| = \sqrt{q\bar{q}}$. One readily checks that the product in $\Bbb H$ satisfies $\overline{q_1q_2}=\bar{q_2}\bar{q_1}$ and $\|q_1q_2\| = \|q_1\|\|q_2\|$. Moreover, if $q\neq 0$, then $q$ is invertible and $q^{-1} = \bar{q}/\|q\|^2$.
The set of purely imaginary quaternions is $\mathrm{Im}(\Bbb H) = \{q \in \Bbb H \mid \mathrm{Re}(q) = 0\} = \mathrm{span}\{i,j,k\} \simeq \Bbb R^3$. Identify $S^3$ with the set of unit length elements in $\Bbb H$. For $u\in S^3$, consider the conjugation by $u$ $$ i_u\colon q \in \mathrm{Im} (\Bbb H) \longmapsto uq\bar{u}\in \mathrm{Im}(\Bbb H) $$ (I leave you as an exercise that $i_u(q)$ is indeed in $\mathrm{Im}(\Bbb H)$ if so is $q$. Hint: show that $i_u$ stabilizes $\Bbb R = \mathrm{Im}(\Bbb H)^{\perp}$, and that it is an orthogonal map). Then $i_u$ is a linear endomorphism of $\mathrm{Im}(\Bbb H)$ satisfying $\|i_u(q)\| = \|q\|$ for all $q$. Moreover, $i_{u_1u_2} = i_{u_1}\circ i_{u_2}$. In other words, we have defined a group homomorphism $$ i \colon S^3 \longrightarrow O(\mathrm{Im}(\Bbb H)) \simeq O_3(\Bbb R). $$ From its definition, $i$ is a continuous map (between Lie groups, which are in particular topological spaces), and $S^3$ is connected. It follows that $i$ has range in the connected component of $id\in O_3(\Bbb R)$, that is, we have a continuous group homomorphism $$ i \colon S^3 \longrightarrow SO_3(\Bbb R). $$ To conclude, it suffices to show:
Indeed, in this case, we would have $SO_3(\Bbb R) \simeq S^3 / \{\pm 1\} = \Bbb RP^3$.
We show the first point. Assume that $i_u = id$ with $u\in S^3$. Then $i_u(q) = q$ for all $q \in \mathrm{Im}(\Bbb H)$ implies that $u$ commutes in particular with $i$, $j$ and $k$. Playing with the anti-commutation properties of $i$, $j$ and $k$ then shows that $u$ has to be real. Hence, $\ker(i) \subset \Bbb R \cap S^3 = \{\pm 1\}$. The reverse inclusion is clear.
Finally, let us show that $i$ is surjective. For $SO_3(\Bbb R)$ is generated by the rotations of angle $\pi$ (that is, orthogonal maps that fix some axis and induce $-id$ in its orthogonal plane), it suffices to show that such maps are attained. Let $P = u^{\perp}$ be a plane in $\Bbb R^3 \simeq \mathrm{Im}(\Bbb H)$, with $u \in \mathrm{Im}(\Bbb H)$ satisfying $\|u\|=1$. From $u^2 = u(-\bar{u}) = -\|u\|^2 = -1$, one has $i_u\circ i_u = i_{u^2} = i_{-1} = id$. Moreover, one checks that $i_u(u) = u$, so that $i_u$ fixes the axis $u^{\perp}$, and hence the plane $u^{\perp}=P$. Ultimately, $u\notin \ker(i)$, so that $(i_u)|_P \neq id_P$, and one thus has $(i_u)|_P = -id_P$. It follows that $i_u$ is the desired rotation of angle $\pi$ in $P$, and that $i$ is surjective.
Remark 1: for surjectivity, we could have done the following instead. Since $S^3$ is compact, so is its image by $i$, which is then closed. By connectedness of $SO_3(\Bbb R)$, it suffices to show that the image of $i$ is open. By group homomorphism properties, it is sufficient to show that $i$ is open near the identity $1 \in S^3$. This latter fact follows from a direct application of the inverse function Theorem. I leave it as an exercise.
Remark 2: we have in fact shown that $SO_3(\Bbb R)$ and $\Bbb R P^3$ are isomorphic as Lie groups.
A nive way to parametrize $S0(3)$ is to consider the map $E: {\bar B}(0,\pi)\to SO(3)$ defined in the following way. Let $x\in {\bar B}(0,\pi)$, and $X$ the linear map $v\to v\wedge x$. Then $X$ is antisymmetric and $E(x)=\exp X$ is the rotation of angle $\vert x \vert$ around the axis oriented by $x$. This map $x\to E(x)$ send the units sphere onto the set of rotation of angle $\pi$ and is exactly 2 to 1 identifying this set with the space $RP^2$. Note that the restriction of $E$ to $B (0,\pi)$ is an homeo onto its image so that $SO(3)= RP^2 \cup B$, $B$ a 3-ball. You see no that $S0(3)$ is obtained by identifying opposite pint of $S^2$ in the 3-ball and is homeomorphic to $RP^3$.
