Zeros of an entire function on the boundary

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Question: Consider a small rectangle as $R_\epsilon=\{x+iy\in\mathbb{C}\mid 1-\epsilon\leq x\leq 1+\epsilon , \ 0\leq y\leq 1+\epsilon\}$ where $\epsilon>0$ is arbitrarily small. For a non zero entire function $f(z)$ which is non-zero on the real axis, prove that there exists $\epsilon_0>0$ such that $f(z)$ has no zeros on the boundary of rectangle $R_\epsilon$ whenever $0<\epsilon<\epsilon_0$.

I tried the following by the method of contradiction: If possible suppose no such $\epsilon_0$ exists. Then for each $\epsilon>0$, there must exist $\epsilon'$ with $0<\epsilon'<\epsilon$ such that the boundary of the rectangle $R_{\epsilon'}$ has atleast one zero of $f$.

In particular if we take $\epsilon_n=\frac{\epsilon_0^*}{2^n}$ where $n\in \mathbb{N}$ and $0<\epsilon_0^*\leq 1$ then there exists $\epsilon_{n}'$ such that $0<\epsilon_n'<\epsilon_n=\frac{\epsilon_0^*}{2^n}$ such that the boundary of $R_{\epsilon_n'}$ has atleast one zero $z_n$ of $f$.

Without loss of generality we may assume $\{\epsilon_n'\}_{n\geq 1}$ is decreasing and $\epsilon_n'\to 0^+$ (To see this choose $\epsilon_1'<\epsilon_1$ and then choose $\epsilon_2'<\epsilon_1'$ and $\epsilon_2'<\frac{\epsilon_0^*}{2^2}$. Next we choose $\epsilon_3'$ such that $\epsilon_3'<\epsilon_2'$ and $\epsilon_3'<\frac{\epsilon_0^*}{2^3}$ and continue this process).

Consider the set $A=\{z_n\in \mathbb{C}\mid n\in \mathbb{N}\}$ where $z_n$ are zeros of $f$ as chosen above. Now we can have two cases:

Case $1$: $A$ is an infinite set- Since $\epsilon_n=\frac{\epsilon_0^*}{2^n}$ so we have $\epsilon_0^*>\epsilon_1>\epsilon_2>\epsilon_3>...$ and hence the zeros $z_n$ of set $A$ lie in a bounded set which is the rectangle $R_{\epsilon_0^*}$ where $0<\epsilon_0^*\leq 1$ which implies that $A$ is a bounded set.

So by the Bolzano-Weierstrass theorem, set $A$ has a limit point. Since $f$ is entire so by the Identity theorem since $A$ is the set of zeros of $f$ and it has a limit point so $f$ must be identically zero which is a contradiction.

Case $2$: $A$ is a finite set- I am having no clue regarding this.

I request you to please give an elaborated answer. Any other method is also welcome.