The Exercise: Let $\phi: A \rightarrow B$ be a ring homomorphism and let $X = \operatorname{Spec} A, Y = \operatorname{Spec} B$. Let $f: Y \rightarrow X$ be the morphism of schemes induced by $\phi$. The exercise states that if 1) $f^{\#}:\mathcal{O}_X \rightarrow f_* \mathcal{O}_Y$ is a surjective morphism of sheaves and 2) $f$ is a homeomorphism onto a closed set of $X$, then $\phi$ must be surjective.
Question: I have written a proof which seems correct but does not use hypothesis 2). So i wonder whether hypothesis 2) is truly necessary.
My Proof: Take $b \in B = f_*\mathcal{O}_Y(X)$. Since $f^{\#}$ is surjective, by exercise II.1.3(a) there exists an open covering $X = \bigcup_i U_i$ and sections $t_i \in \mathcal{O}_{X} (U_i)$ such that $b|_{U_i} = f^{\#}_{U_i}(t_i)$. Since we can cover each $U_i$ be principal opens $D(a_{ij})$, and since $X$ is quasi-compact, we have $X = \bigcup_{i \in I} D(a_i)$, where $I$ is finite, and sections $t_i \in \mathcal{O}_X(D(a_i))$ such that $b|_{D(a_i)} = f^{\#}_{D(a_i)}(t_i)$. Since $ \mathcal{O}_X(D(a_i)) = A_{a_i}$, we have that each $t_i$ is an element of $A_{a_i}$, say $t_i = u_i / a_i^{n_i}$. Since $f_* \mathcal{O}_Y(D(a_i)) = B_{\phi(a_i)}$, and $f^{\#}_{D(a_i)}:A_{a_i} \rightarrow B_{\phi(a_i)}$ is the localization of $\phi$, the equality $b|_{D(a_i)} = f^{\#}_{D(a_i)}(t_i)$ is equivalent to the equality $b/1 = \phi(u_i) / \phi(a_i^{n_i})$ in the ring $B_{\phi(a_i)}$ or equivalently $\phi(a_i^{\ell_i}) b = \phi(a_i^{\nu_i}) \phi(u_i)$ in the ring $B$, for some $\ell_i,\nu_i$. Since the $a_i$ generate $A$, so will the $a_i^{\ell_i}$ and so there exist $c_i \in A$ such that $\sum_i c_i a_i^{\ell_i} = 1$. But this implies that $b = \sum_i \phi(c_i)\phi(a_i^{\nu_i}) \phi(u_i)$ and we are done.
Let $\DeclareMathOperator{Spec}{\operatorname{Spec}}\mathfrak p \in \Spec A$.
We first show that the map $f^\#_\mathfrak p : \mathcal O_{\Spec A, \mathfrak p} \to f_* \mathcal O_{\Spec B, \mathfrak p}$ is the homomorphism $\varphi_\mathfrak p : A_\mathfrak p \to B_\mathfrak p$. Since the basic open subsets $D(g)$ for $g \in A$ form a basis for $\Spec A$, we have \begin{align*} f_* \mathcal O_{\Spec B, \mathfrak p} &= \varinjlim_{\mathfrak p \in D(g)} \mathcal O_{\Spec B}(f^{-1}(D(g))) \\ &= \varinjlim_{\mathfrak p \in D(g)} \mathcal O_{\Spec B}(D(\varphi(g))) \\ &= \varinjlim_{g \not \in \mathfrak p} B_{\varphi(g)} \\ &= B_\mathfrak p. \end{align*}
Since $f^\#$ is induced by $\varphi$, the claim now follows.
If $f^\#$ is surjective, then the induced maps on stalks are also surjective. By proposition 3.9 in Atiyah-MacDonald, it follows that $\varphi$ is surjective.