[proofplan] We prove both directions of the equivalence separately. (⇐) Assume $L$ is the splitting field of some $f \in K[t]$. To show $L/K$ is normal, take any irreducible $g \in K[t]$ that has a root $\alpha \in L$. We must show $g$ splits completely in $L$. Let $\beta$ be any other root of $g$ in $\bar{K}$. The canonical $K$-isomorphism $K(\alpha) \cong K(\beta)$ extends to an embedding $L \hookrightarrow \bar{K}$, but since $L$ is a splitting field the image must be $L$ itself, forcing $\beta \in L$. (⇒) Assume $L/K$ is normal and finite. Write $L = K(\alpha_1, \dots, \alpha_r)$ by the primitive element theorem (or just finiteness of generators). Let $p_i = \min(\alpha_i, K)$. Set $f = \prod p_i$. Each $p_i$ has a root in $L$, so by normality each $p_i$ splits in $L$, hence $f$ splits in $L$. No proper subfield of $L$ can split $f$ because the $\alpha_i$ generate $L$. Thus $L$ is the splitting field of $f$. [/proofplan]

[step:Direction (⇐): Splitting field implies normal]Assume $L$ is the splitting field of some polynomial $f \in K[t]$ over $K$. We show $L/K$ is normal, i.e., every irreducible polynomial in $K[t]$ that has at least one root in $L$ splits completely in $L$. Let $g \in K[t]$ be irreducible with a root $\alpha \in L$, and let $\beta \in \bar{K}$ be any other root of $g$. Since $g$ is irreducible over $K$ and both $\alpha, \beta$ are roots, there is a $K$-isomorphism \begin{align*} \sigma_0 \colon K(\alpha) &\xrightarrow{\;\sim\;} K(\beta) \\ \alpha &\longmapsto \beta. \end{align*} Write $f = c \prod_{i=1}^{n}(t - \gamma_i)$ with each $\gamma_i \in L$, so that $L = K(\gamma_1, \dots, \gamma_n)$. We extend $\sigma_0$ one root at a time. Suppose we have already extended to a $K$-embedding \begin{align*} \sigma_j \colon K(\alpha, \gamma_1, \dots, \gamma_j) \hookrightarrow \bar{K}. \end{align*} The element $\gamma_{j+1}$ is a root of some irreducible factor $h$ of $f$ over $K(\alpha, \gamma_1, \dots, \gamma_j)$. Applying $\sigma_j$ to the coefficients of $h$ gives an irreducible polynomial $\sigma_j(h)$ over $\sigma_j\bigl(K(\alpha, \gamma_1, \dots, \gamma_j)\bigr)$, and we choose any root of $\sigma_j(h)$ in $\bar{K}$ to define $\sigma_{j+1}$. After $n$ steps we obtain a $K$-embedding \begin{align*} \sigma \colon L = K(\alpha, \gamma_1, \dots, \gamma_n) \hookrightarrow \bar{K}. \end{align*} Now observe what $\sigma$ does to $f$. Since $f \in K[t]$ and $\sigma$ fixes $K$, we have $\sigma(f) = f$. Since $f$ splits in $L$ as $f = c\prod(t - \gamma_i)$, applying $\sigma$ gives \begin{align*} f = \sigma(f) = c\prod_{i=1}^{n}\bigl(t - \sigma(\gamma_i)\bigr). \end{align*} Therefore $\sigma$ merely permutes the roots $\{\gamma_1, \dots, \gamma_n\}$, so $\sigma(\gamma_i) \in \{\gamma_1, \dots, \gamma_n\} \subset L$ for each $i$. Since $L = K(\gamma_1, \dots, \gamma_n)$ and $\sigma$ maps each generator into $L$, we conclude $\sigma(L) \subseteq L$. But $\sigma$ is injective and $[L:K]$ is finite, so $\sigma(L) = L$. In particular, $\beta = \sigma(\alpha) = \sigma_0(\alpha) \in \sigma(L) = L$. Since $\beta$ was an arbitrary root of $g$ in $\bar{K}$, the polynomial $g$ splits completely in $L$. As $g$ was an arbitrary irreducible in $K[t]$ with a root in $L$, the extension $L/K$ is normal. $\blacksquare$[/step]

[guided]This is the standard consequence of the first isomorphism theorem for simple extensions: $K(\alpha) \cong K[t]/(g) \cong K(\beta)$, and the composite sends $\alpha \mapsto \beta$ while fixing $K$.[/guided]

[step:Direction (⇒): Normal implies splitting field]Assume $L/K$ is a finite normal extension. Since $L/K$ is finite, there exist finitely many elements $\alpha_1, \dots, \alpha_r \in L$ such that \begin{align*} L = K(\alpha_1, \alpha_2, \dots, \alpha_r). \end{align*} For each $i$, let $p_i = \min(\alpha_i, K) \in K[t]$ be the minimal polynomial of $\alpha_i$ over $K$. Define \begin{align*} f = \prod_{i=1}^{r} p_i \;\in K[t]. \end{align*} We claim $L$ is the splitting field of $f$ over $K$. **$f$ splits in $L$.** Each $p_i$ is irreducible in $K[t]$ and has the root $\alpha_i \in L$. By the definition of normality, every irreducible polynomial in $K[t]$ with a root in $L$ splits completely in $L$. Therefore each $p_i$ splits in $L$, and hence $f = \prod p_i$ splits in $L$. **No proper subfield suffices.** Suppose $K \subseteq M \subseteq L$ and $f$ splits in $M$. Then every root of each $p_i$ lies in $M$. In particular $\alpha_i \in M$ for every $i$, so \begin{align*} L = K(\alpha_1, \dots, \alpha_r) \subseteq M. \end{align*} Combined with $M \subseteq L$, this gives $M = L$. Therefore $L$ is the splitting field of $f$ over $K$. $\blacksquare$[/step]

[guided]This is exactly where the hypothesis of normality is used: normality guarantees that having one root of an irreducible polynomial forces all roots to lie in $L$.[/guided]