An interesting metric space related to Turing machines
Edit: removed hints, posted my solution.

Here is my solution. We're going to pick a reference point $x$ where $f(x) \in L$ and consider the universe from $x$ and $f(x)$'s points of view. It turns out that every "neighborhood" of a point corresponds to a recursive language. So $L$ is a neighborhood around $f(x)$, and there will be some neighborhood around $x$ that maps to it; this neighborhood is a recursive language.

Lemma. In this space, a language is recursive if and only if it is a neighborhood of each of its strings.

Proof. First, fix a recursive language $L$ and let $x \in L$. Let $K$ be the minimal index of a decider for $L$. Then we have that if $y \not \in L$, $s(x,y) \leq K$, so $d(x,y) \geq 1/2^K$. Thus $d(x,y) &lt; 1/2^K$ implies that $y \in L$.

Second, let $x$ be an arbitrary string and fix $\varepsilon &gt; 0$; let $K = \lfloor \log(1/\varepsilon) \rfloor$. Let $L_K = \{y : d(x,y) &lt; \varepsilon\}$; then $L_K = \{y : s(x,y) &gt; K\}$. Then we can write

$$L_K = \{ y : (\forall j=1,\dots,K) |L(T_j) \cap \{x,y\}| \neq 1\}.$$

But $L_K$ is decidable: On input $y$, one may simulate the first $K$ deciders on $x$ and $y$ and accept if and only if each either accepted both or rejected both. $~\square$

Now we're almost done:

Prop. Let $f$ be continuous. If $L$ is recursive, then $f^{-1}(L)$ is recursive.

Proof. Under a continuous function, the preimage of a neighborhood is a neighborhood.



Interestingly, I think that in this space a continuous function is uniformly continuous:  Let $f$ be continuous, so for each point $x$, for each $\varepsilon$ there exists a corresponding $\delta$. Fix an $\varepsilon$ and let $K = \lfloor \log(1/\varepsilon) \rfloor$. There are a finite number of balls of size $\varepsilon$: there is $L(T_1) \cup L(T_2) \dots \cup L(T_K)$; then there is $\overline{L(T_1)} \cup L(T_2) \dots \cup L(T_K)$; then $L(T_1) \cup \overline{L(T_2)} \dots \cup L(T_K)$, and so on. $f$ associates to each of these languages $L_i$ a preimage language $L_i^{\prime}$ with associated diameter $\delta_i$. For each $x \in L_i^{\prime}$, $d(x,y) \leq \delta_i \implies d(f(x),f(y)) \leq \varepsilon$. So we can take the minimum over these finitely many $\delta$s to get the uniform continuity constant $\delta$ associated with this $\varepsilon$.