True or False: Let $\gamma: [0, 1] \rightarrow \{z ∈ C | |z| ≤ 1\}$ be a non-constant continuous mapping such that $\gamma(0) = 0$. Let $f$ be an analytic function in the disc $\{z ∈ C | |z| < 2\}$, such that $f(0) = 0$ and $f(1) = 1$. Then, there exists $\tau$ such that $0 < \tau< 1$ and such that for all $0 < t < \tau$, we have that $f(\gamma(t))\neq 0$.
I think the statement is true. We try to prove it using contradiction method.
If possible assume that for each $0<\tau<1$ we get at least one$0<t<\tau$ such that $f(\gamma(t))=0$. In particular for each $n\in \mathbb N$ we get $0<t_n<\frac{1}{n}$ such that $f(\gamma(t_n))=0$. Now $t_n\rightarrow 0$ as $n \rightarrow \infty $. Since $\gamma$ is a continuous function we get $\gamma(t_n)\rightarrow 0$ as $n\rightarrow \infty$. So we get a sequence $\{ \gamma(t_n) \}$ in $\{z:|z|<2 \}$ which converges in $\{z:|z|<2 \}$ . So by identity theorem $f$ must be a zero function which is not possible as $f(1)=1$. Hence our supposition is wrong.
Am I correct?