= 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 $1,\alpha,\ldots,\alpha^d$ 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.
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