Because the complex numbers are constructed from the real numbers, they are inherently an 'analytic object'. It is thus perhaps not surprising that "there is no" proof that $\mathbb C$ is algebraically closed "without using analysis". Cf. Is there a purely algebraic proof of the Fundamental Theorem of Algebra?
A friend once imagined that, if we take a purely algebraic characterization of $\mathbb C$ as its definition, we may be able to prove that it is algebraically closed "without using analysis".
Specifically, he proposed the following characterization due to Pontryagin: $\mathbb C$ and $\mathbb F_2$ are the only locally compact Hausdorff topological fields with connected group of units. (See Are $\Bbb R$ and $\Bbb C$ the only connected, locally compact fields?)
With $\mathbb C$ defined as the unique topological field different from $\mathbb F_2$ satisfying Pontryagin's characterization, can we give a direct proof that it is algebraically closed?
Ideas: if we assume $F/\mathbb C$ is a finite extension, then the product topology on $F$ does not depend on the choice of a $\mathbb C$-basis and makes $F$ a topological field. With this topology, $F$ satisfies Pontryagin's characterization and so it is isomorphic to $\mathbb C$. If it were true that a field cannot contain itself as a finite-index copy, we would be done. Unfortunately, that is not true (it is not hard to construct a counterexample from a transcendence basis).