$ (\Rightarrow) $ If there is a sequence of odd polynomials $ {p_n(x)} $ with $ p_n \rightarrow f $, then $ f(0) = 0 $. Since $ p_n(x) $ are odd polynomials, then $ p_n(0) = 0. $. Then $ f(0) = \lim_{n \rightarrow \infty} p_n(0) = 0 $
$ (\Leftarrow) $
If $ f \in C([0,1]) $, and $ f(0) = 0 $, then let $ g(x) = f(x) $ if $ x \in [0,1] $, and $ g(x) = - f(-x) $ if $ x \in [-1,0] $. In this case $ g $ is an odd function and $ g \in C([-1,1]) $. Then there exists a sequence of polynomials $ g_n(x) $ such that $ g_n(x) \rightarrow g $ uniformly on [-1,1]. Let $ p_n(x) = \frac{1}{2}(g_n(x)-g_n(-x)) $, then $ p_n(x) $ is an odd polynomial. and $ \lim_{n \rightarrow \infty}p_n(x) = \lim_{n \rightarrow \infty} \frac{1}{2}(g_n(x)-g_n(-x)) = \frac{1}{2}(g(x)-g(-x)) = g(x) $. Then$ p_n(x) \rightarrow f(x) $ on $ [0,1] $.
