1. Graph Theoretic Interpretation: Consider the city as a graph $G$ where houses are vertices and roads are edges. Let $A$ and $B$ be any two vertices (houses) in $G$. According to the problem, for any two vertices $A$ and $B$, there exists exactly one vertex $C$ such that both $A$ and $B$ are connected to $C$.

2. Adjacency and Subgraph: Let $A$ be a vertex in $G$ and let $B_1, B_2, \ldots, B_k$ be the vertices adjacent to $A$. The subgraph induced by the vertices $B_1, B_2, \ldots, B_k$ consists of $\frac{k}{2}$ disjoint edges. This is because if we pick any vertex $X$ in $B$, there must be a vertex $Y$ in $B$ such that $XY$ is an edge. No other vertex in $B$ can be connected to $X$ or $Y$ without contradicting the uniqueness condition.

3. **Vertices Not in $A$ or $B$**: Consider any other vertex $T$ in the graph that is not in $A$ or $B$. By the problem's condition, $T$ must be connected to exactly one vertex in $B$. This is because if $T$ were connected to more than one vertex in $B$, it would contradict the uniqueness condition.

4. Sets of Adjacent Vertices: We can partition the vertices into sets $S_1, S_2, \ldots, S_k$ such that a vertex $X$ is in $S_i$ if and only if $X$ is connected to $B_i$. Without loss of generality, assume $B_1$ is connected to $B_2$, $B_3$ is connected to $B_4$, and so on.

5. Degree Calculation: Pick a vertex $Q$ in $B_1$. For each $i > 2$, $Q$ must be connected to a vertex in $B_i$. This is because the only vertices connected to $B_i$ are in $S_i$. Therefore, $Q$ has at least degree $k$, as it is connected to $B_1$, another vertex in $S_1$, and there are $k-2$ sets $S_{i>2}$.

6. Equality of Degrees: Given any two vertices $A$ and $B$ not connected to each other, we have $\deg(A) \geq \deg(B)$ and $\deg(B) \geq \deg(A)$. This implies $\deg(A) = \deg(B)$.

$\blacksquare$