Solution:

Pick any street $s$ and organize the intersections along $s$ such that the intersections of the two types alternate, as in the statement of the problem.

On every other street $s_{1}$, exactly one intersection has been organized, namely the one where $s_{1}$ intersects $s$. Call this intersection $I_{1}$. We want to organize the intersections along $s_{1}$ such that they alternate between the two types. Note that, as $I_{1}$ is already organized, we have exactly one way to organize the remaining intersections along $s_{1}$.

For every street $s_{1} \neq s$, we can apply the procedure described above. Now, we only need to show that every intersection not on $s$ is well-organized. More precisely, this means that for every two streets $s_{1}, s_{2} \neq s$ intersecting at $s_{1} \cap s_{2}=A$, $s_{1}$ is the main street on $A$ if and only if $s_{2}$ is the side street on $A$.

Consider also the intersections $I_{1}=s_{1} \cap s$ and $I_{2}=s_{2} \cap s$. Now, we will define the "role" of the street $t$ at the intersection $X$ as "main" if this street $t$ is the main street on $X$, and "side" otherwise. We will prove that the roles of $s_{1}$ and $s_{2}$ at $A$ are different.

Consider the path $A \rightarrow I_{1} \rightarrow I_{2} \rightarrow A$. Let the number of intersections between $A$ and $I_{1}$ be $u_{1}$, the number of these between $A$ and $I_{2}$ be $u_{2}$, and the number of these between $I_{1}$ and $I_{2}$ be $v$. Now, if we go from $A$ to $I_{1}$, we will change our role $u_{1}+1$ times, as we will encounter $u_{1}+1$ new intersections. Then, we will change our street from $s_{1}$ to $s$, changing our role once more. Then, on the segment $I_{1} \rightarrow I_{2}$, we have $v+1$ new role changes, and after that one more when we change our street from $s_{1}$ to $s_{2}$. The journey from $I_{2}$ to $A$ will induce $u_{2}+1$ new role changes, so in total we have changed our role $u_{1}+1+1+v+1+1+u_{2}+1=u_{1}+v+u_{2}+5$, As we try to show that roles of $s_{1}$ and $s_{2}$ differ, we need to show that the number of role changes is odd, i.e. that $u_{1}+v+u_{2}+5$ is odd.

Obviously, this claim is equivalent to $2 \mid u_{1}+v+u_{2}$. But $u_{1}, v$ and $u_{2}$ count the number of intersections of the triangle $A I_{1} I_{2}$ with streets other than $s, s_{1}, s_{2}$. Since every street other than $s, s_{1}, s_{2}$ intersects the sides of $A I_{1} I_{2}$ in exactly two points, the total number of intersections is even. As a consequence, $2 \mid u_{1}+v+u_{2}$ as required.