By Lemma \ref{lemma-power-works} we may replace $A$ by $A/(\bigcap \mathfrak p)$ and assume that $A$ is a reduced ring (some details omitted). We may also replace $K$ by the image of $K \to L$. Then $K$ is a reduced ring. The map $\Spec(L) \to \Spec(K)$ is surjective and closed (details omitted). Hence $\Spec(K)$ is a finite discrete space. It follows that $K$ is a finite product of fields. \medskip\noindent Let $\mathfrak p_j$, $j = 1, \ldots, m$ be the minimal primes of $A$. Set $L_j$ be the fraction field of $A_j$ so that $L = \prod_{j = 1, \ldots, m} L_j$. Let $A_j$ be the normalization of $A/\mathfrak p_j$. Then $A_j$ is a semi-local Dedekind domain with at least one maximal ideal, see Algebra, Lemma \ref{algebra-lemma-integral-closure-Dedekind}. Let $n$ be the sum of the numbers of maximal ideals in $A_1, \ldots, A_m$. For such a maximal ideal $\mathfrak m \subset A_j$ we consider the function $$ v_{\mathfrak m} : L \to L_j \to \mathbf{Z} \cup \{\infty\} $$ where the second arrow is the discrete valuation corresponding to the discrete valuation ring $(A_j)_{\mathfrak m}$ extended by mapping $0$ to $\infty$. In this way we obtain $n$ functions $v_1, \ldots, v_n : L \to \mathbf{Z} \cup \{\infty\}$. We will find an element $x \in K$ such that $v_i(x) < 0$ for all $i = 1, \ldots, n$. \medskip\noindent First we claim that for each $i$ there exists an element $x \in K$ with $v_i(x) < 0$. Namely, suppose that $v_i$ corresponds to $\mathfrak m \subset A_j$. If $v_i(x) \geq 0$ for all $x \in K$, then $K$ maps into $(A_j)_{\mathfrak m}$ inside the fraction field $L_j$ of $A_j$. The image of $K$ in $L_j$ is a field over $L_j$ is algebraic by Algebra, Lemma \ref{algebra-lemma-integral-under-field}. Combined we get a contradiction with Algebra, Lemma \ref{algebra-lemma-valuation-ring-cap-field-finite}. \medskip\noindent Suppose we have found an element $x \in K$ such that $v_1(x) < 0, \ldots, v_r(x) < 0$ for some $r < n$. If $v_{r + 1}(x) < 0$, then $x$ works for $r + 1$. If not, then choose some $y \in K$ with $v_{r + 1}(y) < 0$ as is possible by the result of the previous paragraph. After replacing $x$ by $x^n$ for some $n > 0$, we may assume $v_i(x) < v_i(y)$ for $i = 1, \ldots, r$. Then $v_j(x + y) = v_j(x) < 0$ for $j = 1, \ldots, r$ by properties of valuations and similarly $v_{r + 1}(x + y) = v_{r + 1}(y) < 0$. Arguing by induction, we find $x \in K$ with $v_i(x) < 0$ for $i = 1, \ldots, n$. \medskip\noindent In particular, the element $x \in K$ has nonzero projection in each factor of $K$ (recall that $K$ is a finite product of fields and if some component of $x$ was zero, then one of the values $v_i(x)$ would be $\infty$). Hence $x$ is invertible and $x^{-1} \in K$ is an element with $\infty > v_i(x^{-1}) > 0$ for all $i$. It follows from Lemma \ref{lemma-semi-local-dimension-one-conductor} that for some $e < 0$ the element $x^e \in K$ maps to an element of $\mathfrak m_A/\mathfrak p_j \subset A/\mathfrak p_j$ for all $j = 1, \ldots, m$. Observe that the cokernel of the map $\mathfrak m_A \to \prod \mathfrak m_A/\mathfrak p_j$ is annihilated by a power of $\mathfrak m_A$. Hence after replacing $e$ by a more negative $e$, we find an element $a \in \mathfrak m_A$ whose image in $\mathfrak m_A/\mathfrak p_j$ is equal to the image of $x^e$. The pair $(a, x^e)$ satisfies the conclusions of the lemma.