Is any finite-dimensional extension of a field, say $F$, algebraic and finitely generated?

A similar but easier answer to the one given above is as follows.

To show $K/F$ is algebraic if finite we must show that every element of $K$ satisfies a polynomial over $F$.

Suppose $[K : F] = n$ and choose $\alpha\in K$. Then consider the elements $1,\alpha,\alpha^2,...,\alpha^n$.

This is a list of $n+1$ elements in an $n$ dimensional $F$-vector space so must be linearly dependent. Thus there exists $a_0,a_1,...,a_n\in F$ not all zero such that $a_n \alpha^n + ... + a_2\alpha^2 + a_1\alpha + a_0 = 0$.

But then $\alpha$ is a root of the polynomial $a_nx^n + ... + a_2x^2 + a_1x + a_0$ over $F$.

By definition, a field extension of finite degree is finitely generated because the degree is the number of linearly independent "vectors" in the extension required to form a spanning set.

Now let $K/F$ be a finite extension. Consider an arbitrary $a \in K$ and the evaluation homomorphism $\text{ev}_a:F[x] \rightarrow K$ defined such that $g(x) \mapsto g(a)$. This map cannot be injective because $K$ is finitely generated whereas $F[x]$ is not, so the kernel of this map must be a nontrivial ideal. This is to say: we can find nontrivial polynomials in $F[x]$ with $a$ as a root for any $a \in K$, namely the nonzero elements of $\ker(\text{ev}_a)$. In other words, $K/F$ is an algebraic extension.

Other answers provide nice proofs, here is a very short one based on the multiplicativity of the degree over field towers: If $ K/F $ is a finite extension and $ \alpha \in K $, then $ F(\alpha) $ is a subfield of $ K $, and we have a tower of fields $ F \subseteq F(\alpha) \subseteq K $. The Tower Law then asserts that

$$ [F(\alpha):F][K:F(\alpha)] = [K:F] $$

Since $ K/F $ is finite, by definition the RHS is finite. On the other hand, it is seen from the above equality that $ [F(\alpha):F] $ is less than or equal to $ [K:F] $, therefore is also finite. By definition, $ \alpha $ is algebraic over $ F $.