Let \(k\) be an algebraically closed field, let \(V\) be a finite-dimensional \(k\)-vector space, and let \(A\colon V\to V\) be \(k\)-linear. If \(W\subseteq V\) is a nonzero subspace with \(A(W)\subseteq W\), then \(W\) contains an eigenvector of \(A\).
The restriction \(A|_W\colon W\to W\) is a linear operator on a nonzero finite-dimensional space, so its characteristic polynomial has positive degree and has a root \(\lambda\in k\). Then \(A|_W-\lambda\id_W\) is not injective, and a nonzero \(w\in W\) with \(Aw=\lambda w\) is an eigenvector of \(A\).
A linear operator on an infinite-dimensional vector space over an algebraically closed field need not have an eigenvector: multiplication by \(x\) on \(V=W=k[x]\) has no eigenvector, since \(xf=\lambda f\) has no nonzero solution \(f\in k[x]\) by comparing degrees.