Weight space of a representation of ${\frak sl}(2,\mathbb C)$ 
Suppose $(\pi,V)$ is a finite representation of $SU(2)$. Then there's
  an induced representation $(\pi_*,V)$ of the complexified ${\frak su}^\mathbb C(2) = {\frak sl}(2,\mathbb C)$. Show that the weight
  space $\Lambda (\pi_*)$ satisfies
$$ \Lambda (\pi_*) \subset \frac{1}{2}\mathbb Z \alpha $$
where $\alpha$ is one of the roots of  ${\frak sl}(2,\mathbb C)$ .

I know that if $\lambda_1,\lambda_2 \in \Lambda (\pi_*)$ are two weights, then they must differ by an integer of the root:
$$ \lambda_1 - \lambda_2 \in \mathbb Z \alpha $$
This form is very suggestive, but I'm having some trouble applying it without knowing anything about the representation.
Since $\alpha (H) = 2$ for $ H = \left( \begin{array}{ccc}
1 & 0  \\
0 & -1 \end{array} \right) $, I have:
$$ \lambda_1 (H) - \lambda_2 (H) \in \mathbb Z \alpha(H) = 2\mathbb Z $$
So if I knew that there's at least one weight $\lambda_2$ such that $\lambda_2(H)$ is even, then I would deduce that $\lambda_1(H)$ must be even as well (similar if $\lambda_2(H)$ is odd). But I don't know that! 

I'll expand my comment into an answer. 
Note that it suffices by complete reducibility to show that for each irreducible finite dimensional representation of $\mathfrak{sl}_{2}(\mathbb{C})$, the weights are in $\frac{1}{2} \mathbb{Z} \alpha$. Again, by what you mentioned in your question, it suffices to do so for one weight so lets focus on the highest weight of $V$, where we let $\alpha$ be the positive root and $-\alpha$ the negative root.
From here on let, $\mathfrak{g}$ denote $\mathfrak{sl}_{2}.$ Let $E$ be a nonzero element in the root space $\mathfrak{g}_{\alpha}$ and $F$ be a nonzero element in the root space $\mathfrak{g}_{-\alpha}$ such that $[E, F] = H$, where my $H$ is your $H$. 
Basically, 
$$E = \left(\begin{array} {rr}
0 & 1\\
0 & 0 \end{array} \right)$$
and $F$ is its transpose.
Now, let $v$ be a highest weight vector of $V$ (with weight $\omega$.) By definition, a highest weight vector is one that is killed by $E$. Then, for any $n$,
$$F^{n}(v) \in V_{\omega - n\alpha}.$$
By finite-dimensionality, this must stop somewhere i.e. there must be some smallest $n$ such that $V_{\omega - n\alpha} = 0.$
So, we see that $F^{n}v = 0$. Hence, $EF^{n}(v) = 0.$ Now, we use the commutation relations to move everything upwards.
$\begin{align*}
EF^{n}(v) &= [E, F] F^{n-1}(v) + F(EF^{n-1})(v) \\
&= [E, F] F^{n-1}(v) + F [E, F] F^{n-2}(v) + F^{2}EF^{n-2}(v)\\
&= \cdots \\
&= [E, F] F^{n-1}(v) + \cdots F^{n-1}[E, F](v) + F^{n}E(v).
\end{align*}$
Now, we have $E(v) = 0$ because $v$ is a highest weight vector. Additionally, we have $[E, F] = H.$ Finally, we note that $F^{i}(v) \in V_{\omega - i\alpha}$ and hence $H(F^{i}(v)) = (\omega - i \alpha) F^{i}(v).$ Thus, the above equation simplifies to
$$0 = EF^{n}(v) = \left(\sum_{i=0}^{n-1} (\omega - i\alpha)\right) F^{n-1}(v).$$
By assumption on $n$, $F^{n-1}(v) \not = 0$ and hence we have
$$0 = \sum_{i=0}^{n-1}(\omega - i \alpha) = n\omega - \frac{n(n-1)}{2} \alpha.$$
Thus, since $v$ is nonzero and hence $n > 0$, we have $\omega = \frac{n-1}{2} \alpha$, as desired.
