If $L/K$ is a field extension and $K$ is countable, if $[L:K]$ is finite then $L$ is countable

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I have seen from many textbooks and videos that $[\mathbb{R}:\mathbb{Q}]$ is infinite, since $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable. However, I guess I don't fully understand why this is true.

Question:

If $L/K$ is a field extension, and $K$ is countable, then $[L:K]$ being finite implies that $L$ is countable.

What I've tried:

Suppose $[L:K]=d<\infty$.

Then there exists a finite, countable basis $\{b_\alpha\}_{1\leq \alpha \leq d}\subseteq L$ of $L$.

...so what? I'm not sure where to go from here.

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If $d = [L:K]< \infty$. there is a finite basis of $L$ over $K$. We can find $e_1,....,e_d$ in $L$ such that $$L = vect(e_1,...,e_d) = \{x_1 e_1+...+x_d e_d \mid x_1\in K,..., x_d \in K\}$$ So $L$ is in bijection with $K^d = K \times K \times ... \times K$. If $K$ is countable, so is $K^d$.