Continuity and open preimages

$\Leftarrow$. Given $x_0$ and $\varepsilon>0$, the ball $U=B(f(x_0),\varepsilon)$ is open. So $f^{-1}(U)$ is open; it contains $x_0$, hence $B(x_0,\delta)\subseteq f^{-1}(U)$. So $\|f(x)-f(x_0)\|<\varepsilon$ for $\|x-x_0\|<\delta$.

$\Rightarrow$. Given open $U$ and $x_0\in f^{-1}(U)$, $f(x_0)\in U$ gives $\varepsilon>0$ with $B(f(x_0),\varepsilon)\subseteq U$; continuity gives $\delta$ with $f(B(x_0,\delta))\subseteq U$.

Conclusion. Continuity is equivalent to the preimage of open sets being open.