Justification for Freezing Point Depression & Boiling Point Elevation in Solutions?
I think you are looking at the problem from slightly the wrong angle. The central quantity when dealing with colligative properties is entropy and not solute-solvent or solvent-solvent molecule interactions. Of course the interactions are important in the sense that they affect the entropy but since we are dealing with thermodynamics here the way you think about it at the moment clouds the main effect at work here imo.

So, when trying to picture how colligative properties work on a microscopic level think about the effect of the solute on the entropy. Boltzmann's entropy formula tells you that the entropy $S$ is proportional to the logarithm of the number of microstates available to the system. The solute is present in the solution in small quantities and the molecules are really dissolved and don't form a different phase. By their presence the solute molecules disturb the local order of the solvent molecules and introduce new available microstates into the system. As an extremely simplified example you can think of it like this: If you want to place 10 identical blue balls (solvent molecules) on 10 fixed spots there is only 1 way (microstate) to do this, since all the balls are alike. But if you replace one of the blue balls with a red ball (solute molecule) you suddenly have 10 possible distiguishable ways to arrange your balls on the 10 spots. So, when comparing the initial situations of a solution of pure solvent and one of solvent plus solute, the second solution will have a higher entropy to start with, i.e. $S_{\mathrm{l}}^{\mathrm{impure}} &gt; S_{\mathrm{l}}^{\mathrm{pure}}$.

Now, what happens when you want to freeze both solutions? In order to freeze the Gibbs free energy of the frozen (solid) phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{s}} \leq G_{\mathrm{l}}$ or put differently $\Delta_{\mathrm{m}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{l}} - G_{\mathrm{s}} \geq 0$. The Gibbs free energy is given by

\begin{equation}
 G = H - TS
\end{equation}

where $H$ is the enthalpy and $T$ is the temperature, so that

\begin{equation}
 \Delta_{\mathrm{m}} G = (\underbrace{H_{\mathrm{l}} - H_{\mathrm{s}}}_{\Delta_{\mathrm{m}} H}) - T (\underbrace{S_{\mathrm{l}} - S_{\mathrm{s}}}_{\Delta_{\mathrm{m}} S})
\end{equation}

Since there is only very little solute present in the second solution you can assume that $\Delta_{\mathrm{m}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{m}} H^{\mathrm{pure}}$. So, the difference between $\Delta_{\mathrm{m}} G^{\mathrm{impure}}$ and $\Delta_{\mathrm{m}} G^{\mathrm{pure}}$ will mainly be caused by the entropical term. Let's have a look at that: the solid phase is assumed to be pure in both cases, i.e. consisting only of solvent molecules, and thus $S_{\mathrm{s}}$ must be the same in both cases. Furthermore, you generally find that the entropy of a solid phase is lower than the entropy of a liquid phase, i.e. $S_{\mathrm{s}} &lt; S_{\mathrm{l}}$. Above I established that $S_{\mathrm{l}}^{\mathrm{impure}} &gt; S_{\mathrm{l}}^{\mathrm{pure}}$. It follows that $\Delta_{\mathrm{m}} S^{\mathrm{impure}} &gt; \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Putting all this together gives you the following picture: For $\Delta_{\mathrm{m}} G$ to become equal to zero the entropical term $T \Delta_{\mathrm{m}} S$ has to balance $\Delta_{\mathrm{m}} H$, which is assumed to be about the same for the pure-solvent and the solvent-plus-solute case. Since $\Delta_{\mathrm{m}} S^{\mathrm{impure}} &gt; \Delta_{\mathrm{m}} S^{\mathrm{pure}}$ the temperature at which freezing becomes possible must be lower in the solvent-plus-solute case, i.e. adding the solute leads to a freezing point depression.

An analogous argument can be made for the Boiling point elevation. The focal point is again that the solute raises the entropy of the liquid phase, thus affecting the entropical term in the Gibbs free energy equation for the phase change. For the boiling process to happen the Gibbs free energy of the gaseous phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{g}} \leq G_{\mathrm{l}}$ or $\Delta_{\mathrm{vap}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{g}} - G_{\mathrm{l}} \leq 0$.
Requiring the Gibbs free energy of vapourization

\begin{equation}
 \Delta_{\mathrm{vap}} G = (\underbrace{H_{\mathrm{g}} - H_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} H}) - T (\underbrace{S_{\mathrm{g}} - S_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} S})
\end{equation}

to be equal to zero gives

\begin{equation}
 \Delta_{\mathrm{vap}} H = T \Delta_{\mathrm{vap}} S
\end{equation}

Again, we can assume that $\Delta_{\mathrm{vap}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{vap}} H^{\mathrm{pure}}$. Since the entropy of a gaseous phase is generally higher than the entropy of a liquid phase, i.e. $S_{\mathrm{g}} &gt; S_{\mathrm{l}}$, and since $S_{\mathrm{l}}^{\mathrm{impure}} &gt; S_{\mathrm{l}}^{\mathrm{pure}}$, it follows that $\Delta_{\mathrm{vap}} S^{\mathrm{impure}} &lt; \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Thus the boiling temperature must be higher in the solvent-plus-solute case, i.e. adding the solute leads to a boiling point elevation.