For proving this we should know that
$\sum_d(V) =2e \quad \to(1)$
Because one edge consist of two vertices and hence contributes two degrees.
For any Graph $G$
$\sum_d(V)=\sum_d(V_{odd})+Σ_d(V_{even}) \quad \to(2)$
From $(1)$ we can say that $\sum_d(V)$ should be even.
$\sum_d(V_{even})$ will be always even ($\because$ sum of even numbers is always even)
So, for $(2)$ to be true $\sum_d(V_{odd})$ should be even ($\because$ even+odd=odd & even+even=even)
In $\sum_d(V_{odd}),$ every degree is odd.
So, for $\sum_d(V_{odd})$ to be Even,
Number of odd-degree vertices should be even ($\because$ only when we add an even number of odd numbers we get an even number).