Exercise 4.4.11

A fact we’ll use is that $g(A)\subseteq B$ if and only if $A\subseteq g^{-1}(B)$. Which is true since

Fix $x\in R$, we are given that $g^{-1}(V_{𝜖}(x))$ is open, meaning there exists a neighborhood $V_{\delta }(x)$ with $V_{\delta }(x)\subseteq g^{-1}(V_{𝜖}(x))$ implying $g(V_{\delta }(x))\subseteq V_{𝜖}(x)$ and thus $g$ is continuous.

Now suppose $g$ is continuous and let $O\subseteq R$ be an open set. If $x\in g^{-1}(O)$ then $g(x)\in O$ and (since $O$ is open) there exists a neighborhood $V_{𝜖}(g(x))\subseteq O$, now there exists $V_{\delta }(x)$ where $g(V_{\delta }(x))\subseteq V_{𝜖}(g(x))\subseteq O$ so we have $V_{\delta }(x)\subseteq g^{-1}(O)$ and are done.