Lemma

Let U be an open set of X, let V be a connected open set of Y, and let $(x_1, y_1)\in U\times V$. Suppose that the restriction of $s$ to $U\times \{y_1\}$ is continuous. Let $\sigma$ be a continuous section of $f$ over $U\times V$ such that $\sigma (x_1, y_1) =s(x_1, y_1)$. There exists a neighbourhood $U'$ of $x_1$ such that $s=\sigma$ on $U'\times V$. In particular, $s$ is continuous in a neighbourhood of $(x_1, y_1)$.

Since the restriction of $s$ to $U\times \{y_1\}$ is continuous and the mapping $f$ is étale, there exists a neighbourhood $U'$ of $x_1$ such that $s(x, y_1) =\sigma (x, y_1)$ for every $x\in U'($I, p. 34, prop. 11, b)). Let $x\in U'$; since the restriction of $s$ to $\{x\} \times V$ is continuous and the mapping $f$ is étale and separated, it follows from Corollary 1, I, p. 34, that $\sigma (x, y) =s(x, y)$ for every $y\in V$. Thus $s$ and $\sigma$ coincide on $U'\times V$.

We now prove the theorem. First let us prove that the mapping $s$ is continuous in a neighbourhood of every point of $X\times \{y_0\}$. Let $x_0\in$ X; one can choose an open neighbourhood U of $x_0$ in X, a connected open neighbourhood V of $y_0$ in Y, and a continuous section $\sigma : U\times V\rightarrow Z$ of $f$ such that $\sigma (x_0, y_0) =s(x_0, y_0)$. By the above lemma, $s$ is continuous in a neighbourhood of $(x_0, y_0)$.

Let us now prove that the mapping $s$ is continuous at every point of $X\times Y$. For $x_0\in X$, let us denote by $C(x_0)$ the set of points $y\in$ Y such that $s$ is continuous in a neighbourhood of $(x_0, y)$. The set $C(x_0)$ is open in Y by definition, and contains $y_0$ by what precedes.

Let us prove that it is closed in Y. Consider a point $y_1$ of Y adherent to $C(x_0)$, an open neighbourhood U of $x_0$ in X, an open neighbourhood V of $y_1$ in Y, and a continuous section $\sigma : U\times V\rightarrow Z$ of $f$ such that $\sigma (x_0, y_1) =s(x_0, y_1)$. Since the restriction of $s$ to $\{x_0\} \times V$ is continuous and $f$ is étale, there exists an open neighbourhood $V'$ of $y_1$ in Y such that $\sigma (x_0, y) =s(x_0, y)$ for every $y\in V'$ (prop. 11 of I, p. 34). Since Y is locally connected, we may suppose that $V'$ is connected. Since $y_1$ is adherent to $C(x_0)$, the set $C(x_0)\cap V'$ is not empty; let $y_2$ be a point of $C(x_0)\cap V'$. By the lemma applied to $(x_0, y_2)$, there exists a neighbourhood $U'$ of $x_0$ contained in U such that $s=\sigma$ over $U'\times V'$. This proves that $C(x_0)$ is closed, for $U'\times V'$ is a neighbourhood of $(x_0, y_1)$. Since Y is connected, we have $C(x_0) = Y$, and this for every $x_0\in X$, whence the theorem.