Proof of Caratheodory's Theorem (for Convex Sets) using Radon's Lemma

You started off well, especially by assuming $m$ is minimal. This is not a simple proof, and figuring it all out on your own can be very challenging. Instead of trying to use the affine dependence of the points directly, as done in the proof of Radon's theorem, you can instead apply the theorem itself on the set $\{x_1,...,x_m\}$ (since, as you mentioned, $m\geq n+2$). This gives us disjoint $I,J\subseteq [m]$ and equivalent convex combinations: $$\sum_{h\in I}x_h\mu_h=\sum_{h\in J}x_h\mu_h,\sum_{h\in I}\mu_h=\sum_{h\in J}\mu_h=1$$ We can rename the points, and WLOG assume $I=\{1,...,k\},J=\{k+1,...,l\}$, and in addition: $$\frac{\alpha_1}{\mu_1}=\min_{i\in I}{\frac{\alpha_i}{\mu_i}}:=t$$ (here, the coefficients $\alpha_i$ are the same coefficients you defined for the given $y$ in your question). This gives us: $$y=\sum_{i=1}^m\alpha_ix_i=\sum_{i=1}^m\alpha_ix_i-t\sum_{i=1}^k\mu_ix_i+t\sum_{j=k+1}^l\mu_ix_i=$$ $$\sum_{i=1}^k(\alpha_i-t\mu_i)x+\sum_{j=k+1}^l(\alpha_i+t\mu_i)x+\sum_{h=l+1}^m\alpha_ix_i$$ $t\mu_i= \frac{\alpha_1\mu_i}{\mu_1}\leq\frac{\alpha_i\mu_i}{\mu_i}$ and so all of the coefficients in this sum are non-nagative. You can also easily verify that the first coefficient is $0$, and the sum of the coefficients is $1$, and therefore we managed to present $y$ as a convex combination of $m-1$ points, in contradiction to the minimality of $m$.