For $\mathfrak{g}$ a semisimple Lie algebra, $\mathfrak{h}$ a choice of Cartan subalgebra, $\Phi$ the set of roots, there is the integrality property that if $\alpha$ is a root then $c \alpha \in \Phi$ if and only if $c = \pm 1$. I am trying to understand the proof of the 'only if' direction as it appears in my textbook Goodman-Wallach (although Humphreys has a similar proof) (I have also seen the proof here, but I want to do it the way it is presented in my book).
To begin with, $s(\alpha) = \text{span}\{e_\alpha, f_\alpha, h_\alpha\} \cong \mathfrak{sl}_2$ is an isomorphic copy inside $\mathfrak{g}$. Construct $$ M_\alpha = \mathbb{C}h_\alpha + \sum_{c \neq 0} \mathfrak{g}_{c\alpha}.$$ Then the eigenvalues of $h_\alpha$ on $M_\alpha$ (under the adjoint representation) are 0 and $2c$. Since $M_\alpha$ is completely reducible and the weights of $\mathfrak{sl}_2$-irreps are known, we conclude that every $2c$ must be an integer, so $c$ is an integral multiple of 1/2. By the classification of $\mathfrak{sl}_2$-irreps (eigenvalues are integers which differ by 2), this then tells us that the set $\{ 2c : \mathfrak{g}_{c\alpha} \in M_\alpha\} $ either consists of all even or all odd integers. In particular, there are two cases: either every $c$ is an integer, or else every $c$ is of the form $p+1/2$ for some positive integer $p$ (they are very similar, so I am posting them as one question).
In the first case, my textbook says that since $s(\alpha)$ contains the zero eigenspace in $M_\alpha$ (I agree with this), then it follows that $c\alpha$ is not a root for any integer $c$ for which $\vert c \vert > 1$. I don't see how this conclusion follows at all. I understand that $s(\alpha)$ is one such irreducible component of $M_\alpha$, and that its eigenvalues are $0, \pm 2$. I also recognise that this exhausts the occurrences of the zero weight, since 0 has multiplicity one in $M_\alpha$. However I don't understand how either of these statements imply the conclusion.
In the second case, if $(p+1/2)\alpha \in \Phi$, then the eigenvalues of $h_\alpha$ are $2p+1, 2p-1, \dots, 3, 1$ on $M_\alpha$, and then somehow this implies $(1/2)\alpha$ is also a root. I also don't see how this follows, although maybe if my first question is answered then this will also follow.
Your agrument goes wrong where you say "... by the classification of $\mathfrak{sl}_2$-irreps ...". What you can conclude there is that $M_\alpha$ is a direct sum of finitely many non-trivial irreducible representations. Now consider irreducible components of odd dimension. On these, $h_\alpha$ has only even eigenvalues and always has a one-dimensional zero-eigenspace. Since the zero-eigenspace in $M_\alpha$ is one-dimensional and $s(\alpha)\subset M_\alpha$ is one irreducible component, you conclude that there may not be any other odd dimensional irreducible components. Hence in all other irreducible components $h_\alpha$ has only odd eigenvalues. In particular, the eigenspace for the eigenvalue $4$ in $M_\alpha$ has to be trivial, so $\mathfrak g_{2\alpha}=\{0\}$ and hence $2\alpha$ cannot be a root.
But now if there were any even dimensional irreducbible components, each of them would have a non-trivial eigenspace for the eigenvalue $1$. But this would imply that $\mathfrak g_{\alpha/2}$ would be non-zero and hence $\alpha/2$ would be a root. But this is a contradiction, since from above, this would imply that $2(\alpha/2)=\alpha$ is not a root.