Answers to the OP's question depend on where the OP is willing to start. To prove that an abstractly defined group is (i.e., has the structure of) a Lie group, one will have to use something nontrivial, as this is not a trivial task, in general.
For example, É. Cartan's statement that the set of (smooth?, $C^k$?, analytic?) diffeomorphims of a manifold $M$ that preserve a (smooth? $C^k$?, analytic?) parallelization is (i.e., has a natural structure as) a Lie group, was taken to be obvious by him, but was not actually proved to modern standards of rigor until much later. It was the basis of all of his results about the group of symmetries of various geometric structures (including conformal geometry in dimensions at least $3$) being Lie groups. In particular, it implies that the group of self-isometries of a pseudo-Riemannian manifold is a Lie group.
If the OP is willing to assume that the group of self-isometries of a Riemannian manifold is a Lie group, then one can give an argument that the group of conformal symmetries of a manifold of dimension at least $3$ is a Lie group as well using only this, and, moreover, one can immediately see, based on Riemannian geometry, where the assumption that the dimension is at least 3 enters the picture. The idea is as follows:
If $g$ is a metric on an $n$-manifold $M$ and $u$ is a smooth function on $M$, then consider the expression $\mathrm{Ric}(e^{2u}g)$, that computes the Ricci curvature of the conformal metric $g' = e^{2u}g$:
$$
\mathrm{Ric}(e^{2u}g) = \mathrm{Ric}(g)-(n{-}2)\left(\mathrm{d}u^2+|\nabla u|^2g\right) - (n{-}2)\,\mathrm{Hess}(u) + (\Delta u)\,g,
$$
where $\mathrm{Hess}(u) = \nabla(\mathrm{d}u)$ is a quadratic form that expresses the second covariant derivatives of $u$ and whose trace is $-\Delta u$.
Thus, when $n\ge 3$, if one specifies the $1$-jet of $u$ at $p$, there will be a unique extension to a $2$-jet of $u$ at $p$ such that $\mathrm{Ric}(e^{2u}g)(p)=0$. (By contrast, when $n=2$, $\mathrm{Ric}(e^{2u}g) = (2K(g)+\Delta u)\,g$, where $K(g)$ is the Gauss curvature of $g$, so specifying $\mathrm{Ric}(e^{2u}g)(p)=0$ only detemines one coefficient of the $2$-jet of $u$ at $p$ instead of all $3$ coefficients.)
This observation can be used to show that, when $n\ge 3$, if we let $C\to M$ be the $\mathbb{R}^+$-bundle of all multiples of $g$, which depends only on the conformal class of $g$, then, on $\pi:J^1C\to M$, the bundle of $1$-jets of sections of $C$, there is a canonical $n$-plane bundle $D\subset T(J^1C)$ that is transverse to the fibers of $\pi:J^1C\to M$ and is such that the $1$-jet graph of a conformal metric $e^{2u}g$ is tangent to the $n$-plane bundle $D$ over a point $p\in M$ if and only if $\mathrm{Ric}(e^{2u}g)(p)=0$. Any conformal transformation of $\bigl(M,[g]\bigr)$ canonically induces a transformation of $J^1C$ that preserves this $n$-plane field $D$. This defines a canonical splitting $T(J^1C) = D\oplus \mathrm{ker}(\pi')$ that can then be used to construct a canonical metric $h$ on $J^1C$ that is preserved by any such transformation. Thus, the group of $[g]$-conformal transformations on $M$ is embedded in the group of self-isometries of $(J^1C,h)$ and hence is embedded as a (closed, it turns out) subgroup of a Lie group. Thus, in this way, it inherits the structure of a Lie group.
By the way, for entirely different reasons, the group of conformal transformations in dimension $2$ is also a Lie group, but it cannot be proved by the method above. Instead, one uses the Uniformization Theorem, which shows that any simply-connected Riemannian $2$-manifold is conformally equivalent to either the disk, the plane, or the sphere (with their standard conformal structures) and then appeals to elementary complex analysis to show that their conformal symmetries are Lie groups.
Finally, I cannot resist pointing out that the group of conformal transformations in dimension $1$ is, nowadays, not regarded as a Lie group, although Lie (and Cartan) regarded it as a Lie group, just infinite dimensional.
