 Solution

An element is algebraic over $F$ if it is a zero of some nonzero polynomial in $F[x]$. An extension is algebraic if every element is algebraic. If $[K:F]=d<\infty$, the $d+1$ elements $1,\alpha ,\dots ,{\alpha }^d$ are linearly dependent over $F$, proving every $\alpha \in K$ is algebraic.

The converse fails for the algebraic closure $\overline{\mathbb{Q}}$ inside $\mathbb{C}$. All its elements are algebraic by definition, but it contains $2^{1/n}$ of degree $n$ for every positive integer $n$, since $x^n-2$ is irreducible by Eisenstein's criterion at two. A finite extension could not contain elements of unbounded degree, so this algebraic extension is infinite.