if the inverse images of all closed balls are closed, is $f$ continuous?

I don't think this is true. What if we define $f: D:= [0, \infty) \rightarrow \mathbb{R}$ by the formula $f(x) = \frac{1}{x}$ for $x > 0$, and $f(0) = 3$. This function is not continuous, because the sequence $\frac{1}{n}$ converges to $0$, but $f(\frac{1}{n})$ does not converge to $f(0) = 3$. Another way of seeing this is that the preimage of the closed set $[4, \infty)$ under $f$ is $(0,\frac{1}{4}]$, which is not closed in $D$.

Now, let's check that the preimage of any closed ball $[a,b]$ in $\mathbb{R}$ is closed in $D$. Let $g$ be the restriction of the function $f$ to the open interval $(0,\infty)$. Note that $g^{-1}[a,b]$ is closed not only in $(0,\infty)$, but also in $D$.

If $[a,b]$ does not contain $3$, then $f^{-1}[a,b]$ is a subset of $(0,\infty)$, and so $f^{-1}[a,b] = g^{-1}[a,b]$ which is closed in $D$. If $3 \in [a,b]$, then $f^{-1}[a,b] = \{0\} \cup g^{-1}[a,b]$, with $g^{-1}[a,b]$ closed in $D$.

Here is the "universal" counterexample. Let $D=\mathbb{R}^n$, with the topology which has as a subbasis the sets $\mathbb{R}^n\setminus B$ for all closed balls $B$. Then the identity map $i:D\to \mathbb{R}^n$ satisfies your condition, but it is not continuous (because, for instance, any nonempty open subset of $D$ contains the complement of a finite union of balls and hence is unbounded). This is universal in that map $f:X\to\mathbb{R}^n$ satisfies your condition iff there is a (necessarily unique) continuous map $g:X\to D$ such that $f=ig$ (namely, $g$ is $f$ considered as a map to $D$).

On the other hand, there are no counterexamples if you require the function to be bounded. To show this, we want to show that $i$ is continuous when restricted to any bounded set. Concretely, this means that if $A\subset\mathbb{R}^n$ bounded, $U\subseteq A$ is open in the usual topology, and $x\in U$, then we can find finitely many closed balls $B_1,\dots, B_m$ such that $x\in A\setminus(B_1\cup\dots B_m)\subseteq U$. To prove this, note that we may assume $A$ is closed and hence compact, so $A\setminus U$ is also compact. For each $y\in A\setminus U$, you can choose an open ball around $y$ whose closure does not contain $x$. Finitely many of these open balls then cover $A\setminus U$ by compactness, and you can take $B_1,\dots,B_m$ to be their closures.