Forgive my lack of rigor because I am professionally not into science/math.
Is it possible to find irreducible linear(in contrast to projective) representations of $SO(3)$ without passing to lie algebras? I want this because Lie algebras have more structure than lie groups and I wanted to know whether it is impossible to find out the irreducible representations of a Lie group without using its lie algebra although I know there is a direct correspondence between them. Exactly how much can be said about a Lie group without bothering about its algebra would be a much broader question, which I am ultimately interested in.
To find out the irreps of $SO(3)$ using the power of Lie algebras I would just evaluate the irreps of the lie algebra which also happens to be the algebra of its universal covering group $Spin(3)=SU(2)$.Then I would exponentiate to obtain the $SU(2)$ Lie group elements and reject the spinorial representations on grounds that they don't obey $R(0)=R(2 n \pi)~~\forall n \in \mathbb{Z}$ which is a necessary requirement to qualify as a "rotation" or a $SO(3)$ element.
But as you can see this derivation requires a lot of (nevertheless "standard") machinery like "vector space", "exponential map" etc. I was wondering whether there is a more "direct" classification of irreducible representations of $SO(3)$ group elements or not.
Edit (in response to comments): I need to show that the construction finds all of them. I am not assuming any knowledge of $SU(2)$ group representation classification because I don't know if I can do that without the algebra.
Yes, this can be done (without using $SU(2)$, even) although proving that you've constructed all the irreducibles is a little tedious. Here is a sketch. First, $SO(3)$ naturally acts on the vector space $\mathbb{R}[x, y, z]$ on homogeneous polynomials in three variables via linear changes of variables. The Laplacian acts on homogeneous polynomials and is $SO(3)$-equivariant, so for each $k$ the subspace of homogeneous harmonic polynomials (polynomials satisfying $\Delta f = 0$) of degree $k$ is $SO(3)$-invariant.
This turns out to be the unique irreducible representation of dimension $2k+1$, and there aren't any others. You can prove this by proving that the characters of these representations separate conjugacy classes, in the sense that if $g, h \in SO(3)$ are not conjugate then their character for at least one of these irreducibles is different. But like I said it's a little tedious.
Edit: To say this last bit in more detail, we have the following corollary of the Peter-Weyl theorem: if $G$ is a compact group then the characters of the finite-dimensional irreducible representations of $G$ form an orthonormal basis of the space of $L^2$ class functions on $G$. So, if you have the characters of some of the irreducible representations of $G$, and you can show that they separate conjugacy classes, then by Stone-Weierstrass they span a dense (in the uniform norm) subspace of class functions on $G$. From this it should follow that they produce an orthonormal basis of $L^2$ class functions. The character of any other irreducible would have to be orthogonal to all of them but since they are a basis there are no such characters, so you've found all the irreducibles.
There should be other easier ways to do this too. You can check out this old blog post where I classify the irreducible representations of $SU(2)$ in $4$ different ways, only one of which involves passing to the Lie algebra. These proofs directly give the desired result for $SO(3)$ once we know about the double cover $SU(2) \to SO(3)$, and I think each of the methods can also be adapted to this case.