Assume $I=[a,b]$ and $f:[a,b] \to \mathbb R$ is continuous on $I$ and for all $x$ in $I$ exists $y \in I$ such that $\left|f\left(y\right)\right|\le\frac{1}{2}\left|f\left(x\right)\right|$ prove that there exists a number $c$ in $I$ such that $f(c)=0$.

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For $x\_1 \in I$ exists $x\_2 \in I$ such that $\left|f\left(x\_2\right)\right|\le\frac{1}{2}\left|f\left(x\_1\right)\right|$,inductively it's possible to find a sequence $(x\_n) \in I$ such that for all $n \in \mathbb N$:$$\left|f\left(x\_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x\_n\right)\right|$$
And so
$$\left|f\left(x\_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x\_{n}\right)\right|\le\left(\frac{1}{2}\right)^{2}\left|f\left(x\_{n-1}\right)\right|\le...\le\left(\frac{1}{2}\right)^{n-1}\left|f\left(x\_{2}\right)\right|\le\left(\frac{1}{2}\right)^{n}\left|f\left(x\_{1}\right)\right|$$

From here we see that the sequence $\left\{f\left(x\_{n}\right)\right\}\_n$ is a bounded sequence in $\mathbb R$ and by Bolzano–Weierstrass theorem there exists a subsequence $\left\{f\left(x\_{n\_k}\right)\right\}\_k$ of $\left\{f\left(x\_{n}\right)\right\}\_n$ which converges to $f(c)$,on the other hand $$f(c)=\lim f(x\_{n\_k})=0 $$

I have two problems here,first how do we really "inductively" can find a sequence $(x\_n) \in I$ such that for all $n \in \mathbb N$:$\left|f\left(x\_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x\_n\right)\right|$

Besides I have not used the fact that $f$ is continuous ,if I could show that $(x\_n)$ is bounded then I could use Bolzano–Weierstrass theorem to conclude that there exists a subsequence $\left(x\_{n\_k}\right)$ which converges to $c$,and continuity of $f$ implies that $f(c)=\lim f(x\_{n\_k})=0 $,but I cannot show that $(x\_n)$ is bounded.