By (Stacks Project, Tag 04GG), R is Henselian if and only if for any étale ring map R \to S and prime \mathfrak{q} of S lying over \mathfrak{m} with \kappa = \kappa(\mathfrak{q}), there exists a retraction S \to R of R \to S.

Suppose R is Henselian and let R \to S \to \kappa be a factorisation of R \to \kappa with R \to S étale. Then \mathfrak{q} = \mathrm{ker}(S \to \kappa) is a prime ideal of S lying above \mathfrak{m}. The map \kappa \to \kappa(\mathfrak{q}) induced by S \to \kappa postcomposes with \kappa(\mathfrak{q}) \to \kappa to the identity of \kappa. Hence \kappa(\mathfrak{q}) = \kappa and by assumption, R \to S has a retraction.

Now assume that R \to \kappa is étale-Henselian and let R \to S be étale and \mathfrak{q} be a prime ideal of S with \kappa(\mathfrak{q}) = \kappa. Hence the composition R \to S \to \kappa(\mathfrak{q}) = \kappa is a factorisation of R \to \kappa. Thus R \to S has a retraction.

Since B/\m B is an étale algebra over the residue field k = A/\m, it splits as B/\m B \cong k_1 \times \cdots \times k_n for finitely many finite separable extensions k_i of k. Thus B/\n, as a quotient of B/\m B, is also a finite separable extension of k. Choose an inclusion B/\n \hookrightarrow k^{\sep}. The composition B \to B/\n \to k^{\sep} then factors A \to k^{\sep} through B. Since A \to k^{\sep} is étale-Henselian, there exists a retraction r : B \to A of A \to B. But this retraction may not satisfy \n = r^{-1}(\m).

To solve this, notice that there are only finitely many prime ideals \n_0 = \n, \n_1, \cdots, \n_k of B lying over \m and they are all maximal ideals (directly by the fact that B/\m B \cong k_1 \times \cdots \times k_n). Since \bigcap_{i \ne 0} \n_i \nsubseteq \n (otherwise one of \n_i \subseteq \n which is impossible), we can find an element b \in \n_i for every i \ne 0 that does not fall in \n. We may replace (B, \n) by (B_b, \n B_b), then B is still étale over A and \n is the only prime ideal of B lying over \m. Then the section s_b : B_b \to A given by the previous paragraph satisfies \n B_b = s_b^{-1}(\m). Composition with the localization map B \to B_b gives the desired section s : B \to A.

Let s : B \to A be the section given by Proposition 1.3.5 for the pair (B, \n). Localizing at \m, we obtain maps f_\m \colon A = A_\m \to B_\m and s_\m \colon B_\m \to A, and we write f' \colon A \to B_\n for the composition of f_\m with the localization B_\m \to B_\n. A map s' \colon B_\n \to A with s' \circ (B_\m \to B_\n) = s_\m exists by the universal property of localization, since \n = s^{-1}(\m). Moreover, s' is a section of f'.

Since f_\m is étale and s_\m is a section of f_\m, it follows that s_\m is also étale by the cancellation property of étale maps. Thus s', being a composition of a localization and an étale map, is flat. As a flat local ring homomorphism, s' is faithfully flat and hence injective. Since s' is also a section of f', it is surjective. Therefore, s' is an isomorphism, and the composition A \to B_\n is an isomorphism.