I'm not sure what theorems you've seen, but if you know the open mapping theorem, then you know that $f(\mathbb C)$ is open. Let $(z_n)$ be a sequence in $f(\mathbb C)$ converging to $z\in\mathbb C$. Then there is a sequence $(w_n)$ with $f(w_n)=z_n$. Since convergent sequences are bounded, $(z_n)$ is a bounded sequence. Since the sequence $(w_n)$ lies in $f^{-1}(\{z_n\})$, $(w_n)$ is also a bounded sequence, and therefore there is a subsequence $(w_{n_k})_k$ converging to some $w\in\mathbb C$. By continuity of $f$, $f(w)=\lim_k f(w_{n_k})=\lim_k z_{n_k}=z$. Therefore $f(\mathbb C)$ is closed. Since $\mathbb C$ is connected, it follows that $f$ is onto.
If $f$ is a nonconstant polynomial, then $\lim_{z\to\infty}f(z)=\infty$, which implies that $f$ satisfies the hypothesis. Therefore $f$ is onto, and in particular $0$ is in its image, which is the fundamental theorem of algebra.
An alternative solution to the original problem that is probably overkill and actually uses the fundamental theorem of algebra goes as follows. If $f$ is a nonconstant polynomial, then it is onto by the fundamental theorem of algebra. If $f$ is an entire function that is not a polynomial, then it has an essential singularity at $\infty$, and hence $f(\{z:|z|>M\})$ is dense in $\mathbb C$ for all $M>0$ by the Casorati-Weierstrass theorem. In particular, $f^{-1}(\{z:|z|<1\})$ intersects $\{z:|z|>M\}$ for all $M>0$, and therefore it is unbounded. (I guess this is a good approach if for some reason you want to avoid using the open mapping theorem.)
Edit: Removed remark about assuming that $f$ is nonconstant, since that is implied by the hypothesis as Theo pointed out.