The circle bundle of $S^2$ and real projective space

$SO_3$ is the space of triples $(v_1,v_2,v_3)$ of elements of $\mathbb R^3$ which are an oriented orthonormal basis.

Given an element $x\in SS^2$, you construct a pair of orthonormal basis for it. $v_1$ is the point of $S^2$ your vector $x$ is tangent to, $x \in T_{v_1}S^2$ and $v_2$ would be the vector in $\mathbb R^3$ that is the image of $x \in T_{v_1}S^2$ under the inclusion of vector-spaces $T_{v_1}S^2 \subset \mathbb R^3$.

But given $v_1$ and $v_2$ orthonormal, $v_3 = v_1 \times v_2$.

So that's essentially why $SO_3$ and $SS^2$ are diffeomorphic / homeomorphic.

There's a lot of fun ways to see $\mathbb RP^3$ and $SO_3$ are diffeomorphic. There are arguments using the quaternions. I prefer the exponential map $T_ISO_3 \to SO_3$ -- consider it restricted to balls of various radius and stop at the first radius where the function is onto.

Here's another approach. I think of it as less elementary but it fits into a fairly broad framework of general-nonsense about bundles.

Fact: The Hopf fibration $S^1 \to S^3 \to S^2$ is the circle bundle over $S^2$ with Euler class $+1$. One way to take this is this bundle is classified by a map $S^2 \to B(SO_2)$, where the induced map $H_2(S^2) \to H_2(B(SO_2))$ sends the generator to the generator. Another way to say this is you can decompose $S^3$ into two solid tori $S^1 \times D^2$, and the gluing map sends the $\{1\} \times S^1$ curve in $\partial (S^1 \times D^2)$ to the diagonal $\{(x,x) : x \in S^1\}$ curve in the boundary of the other $S^1 \times D^2$ i.e. the curve of "slope $1$".

$S^3$ is the group of unit quaternions, so the unit complex numbers $S^1$ is a subgroup. Let $\mathbb Z_n \subset S^1$ be the $n$-th roots of unity. Since $S^3/S^1 \simeq S^2$ by the Hopf fibration, there are also bundles, induced by the Hopf fibration:

$$S^1 / \mathbb Z_n \to S^3 / \mathbb Z_n \to S^2 $$

But $S^1 / \mathbb Z_n$ is a circle. So general bundle nonsense says $S^3 / \mathbb Z_n$ is the circle bundle over $S^2$ with Euler class $n$. In the case $n=2$ this is $\mathbb RP^3$, in general this space is called the Lens space $L_{n,1}$. In particular, the unit tangent bundle of $S^2$ is known (by Poincare-Hopf, for example) to be the circle bundle over $S^2$ with Euler class $2$.

Here's is the outline of the argument using quaternions which Ryan alluded to in his answer.

The quaternions are, as a vector space, isomorphic to $\mathbb{R}^4$. They are the set of all things of the form $q=a + bi + cj + dk$ with $a,b,c$, and $d$ real numbers together with the multiplication rules that $i^2 = j^2 = k^2 = ijk = -1$ and $ij = -ji =k$, $jk = -kj = i$, and $ki = -ik = j$. One can check that this notion of multiplication is associative with unit $1 + 0i + 0j + 0k = 1$ and every nonzero element has a multiplicative inverse. If one defines the norm squared $|q|^2$ of a quaternion as $a^2+b^2+c^2+d^2$, then a computation shows that for any two quaterions $q_1$ and $q_2$, we have $|q_1q_2| = |q_1||q_2|$. It follows that the unit sphere in $\mathbb{R}^4$ has the structure of a (noncommutative) Lie group.

Let $V$ be the subset of all quaternions consisting of those with $a=0$,the imaginary quaternions. Then $V$ is isomorphic to $\mathbb{R}^3$ as a vector space.

For each $q\in S^3$ (i.e, $q$ is a unit length quaternion), define the map $A_q:V\rightarrow V$ which sends $v$ to $qv\overline{q}$ (where $\overline{q}$ negates the $i,j,$ and $k$ terms, but leaves the real part alone). Because $|qv\overline{q}| = |q||v||q| = |v|$, $A_q\in SO(3)$.

That means that $q\rightarrow A_q$ is really a homomorphism from $S^3\rightarrow SO(3)$. Now one shows the following: Every element of $SO(3)$ is in the image of $S^3$ under this map and $1$ and $-1\in S^3$ together make up the whole kernel of the map. It follows that $SO(3)$ is diffeomorphic to $S^3/$~ where $q$~$-1*q$. But this is precisely the identification done on $S^3$ to get $\mathbb{R}P^3$ so we must have that $SO(3)$ and $\mathbb{R}P^3$ are diffeomorphic.