Nash Equilibrium for n-shops Location Game
The following two claims hold in the general $n$-shop case.
Claim 1. In equilibrium a shop closest to an edge (0 or 1) cannot be alone.
Proof. Such a shop could gain customers by moving slightly inward.
Claim 2. In equilibrium at most two shops can be in any location.
Proof. Assume there is an equilibrium where there are three or more shops in a location. Denote the number of customers coming to this location from the left and right by $c_l$ and $c_r$ respectively. (In the 3-shop case $c_l + c_r = 1$.) All shops in the location have a payoff of $(c_l + c_r)/n$. By moving slightly to the left, a shop can attain a payoff that is arbitrarily close to $c_l$, and by moving slightly to the right a shop can attain a payoff that is arbitrarily close to $c_r$, therefore it can attain (almost) $\max(c_l;c_r)$. It is easy to show that if $n &gt;2$ then
$$
\frac{c_l + c_r}{n} &lt; \max(c_l;c_r). 
$$

Using these two properties it is very easy to show that in the 3-shop case there is no pure Nash-equilibrium, so this is left as an exercise.
I happen to have a Desmos for this, if you'd like you can use it to validate the claims and reason about the proof.

Note: a mixed equilibrium does exist, see Shaked, A. (1982): Existence and Computation of Mixed Strategy Nash Equilibrium for 3-Firms Location Problem.

Interestingly, $n = 3$ is the only case without a pure equilibrium, $n \in \left\{2,4,5\right\}$ all have one, and for other $n$ values there are infinitely many equilibria. For a formal characterization one needs some additional properties, but no advanced mathematics. For a detailed discussion see
Eaton, B.C., and R.G. Lipsey (1975): The Principle of Minimum Differentiation Reconsidered: Some New Developments in the Theory of Spatial Competition.