Let $f: X \rightarrow Y$ be a flat morphism of finite $K$-schemes for some field $K$. Let $V\subset X$ be an irreducible component of $X$. I want to know that $W=\overline{f(V)}\subset Y$ is an irreducible component of $Y$, and here's how I think it can be proven. $W$ should be irreducible for topological reasons, so we need only show maximality.
Since $V\subset X$ is an irreducible component, we know that locally it corresponds to a prime ideal sheaf which is a minimal prime, so that $\mathcal{O}_{V,X} = A$ is an Artinian local $K$-algebra. Further, since $f$ is flat, the map $f: \mathcal{O}_{W,Y}=B \rightarrow A$ is a flat local homomorphism of local $K$-algebras, and consequently the induced map $\phi: \operatorname{Spec}A \rightarrow \operatorname{Spec} B$ is surjective. Consequently, as $\operatorname{Spec} A$ is a singleton, $\operatorname{Spec}B$ is also a singleton and so B is Artinian. Since $B$ is the local ring of a closed irreducible subscheme, this should imply that $W$ is a maximal irredicuble subscheme and is hence an irreducible component.
Does this argument work? How can I make more rigorous the idea that $W$ should be maximal?
Edit: @Roland has provided a nice geometric proof of this fact in the comments but I would prefer to know if this works, or if not, a working algebraic proof of the statement.