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Edit: the answer suggested by @ShakedBader posted here works for this question as well. But I'm curious about the extra assumption here about $|f(0)|$. Does it allow to solve the question using Rouche's lemma?

Let $f$ be a holomorphic function on a domain which contains the unit circle.

We know that $|f(z)|>1$ for every $|z|=1$, and that $|f(0)|<1$. Show that there exists a point $a\in \mathbb C$, such that $|a|<1$ and $f(a)=a$.

I see that due to continuity, we can find a path in the unit circle which circles zero, and where $|f(z)|=1$ hold for each $z$. But I'm not sure how to proceed. Any clues?

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    Can't you use Rouche's theorem to show that f(z) and f(z)-z have the same number of zeros on the interior of the disk ? – 2017-02-14
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    Possible duplicate of [Entire function such that $|f(z)| \geq 11|z|$ on $\mathbb {C}-closure({\mathbb {D}(0, 10))}$](http://math.stackexchange.com/questions/2143903/entire-function-such-that-fz-geq-11z-on-mathbb-c-closure-mathbb) – 2017-02-14

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The condition that $|f(0)|<1$ while $|f(z)|>1$ on the unit circle allows you to conclude by continuity on the closed unit disk that $|f(z)|$ must have a local minimum in the open unit disk. This minimum must be $0$, or else you would derive a contradiction from the maximum modulus theorem applied to $\dfrac1f$. Thus, you know that $f$ has a zero inside the disk.

You can combine this with hjr's suggestion to use Rouché's theorem.