Any two natural transformations between identity functors commute

Let $\mathcal{C}$ be a category, $id_\mathcal{C}:\mathcal{C} \to \mathcal{C}$ the identity functor. Prove that for any two natural transformations $\alpha, \beta : id_\mathcal{C} \Rightarrow id_\mathcal{C}$, $$\alpha \circ \beta = \beta \circ \alpha.$$

Here is my solution:

Note that (is this step correct?) $\; \forall f:c \to d$ in $\mathcal{C}$, $$f \circ (\alpha_{c} \circ \beta_{c}) = (\beta_{d} \circ \alpha_{d}) \circ f, \quad f \circ (\beta_{c} \circ \alpha_{c}) = (\alpha_{d} \circ \beta_{d}) \circ f.$$

Since $\alpha \circ \beta$ and $\beta \circ \alpha$ are natural transformations (via vertical compositions), we have $$f \circ (\alpha_{c} \circ \beta_{c}) = (\alpha_{d} \circ \beta_{d}) \circ f, \quad f \circ (\beta_{c} \circ \alpha_{c}) = (\beta_{d} \circ \alpha_{d}) \circ f.$$ Therefore, $$f \circ (\alpha_{c} \circ \beta_{c}) = f \circ (\beta_{c} \circ \alpha_{c}), \quad (\alpha_{d} \circ \beta_{d}) \circ f = (\beta_{d} \circ \alpha_{d}) \circ f.$$ It follows that (is this step correct?) $$\alpha_{c} \circ \beta_{c} = \beta_{c} \circ \alpha_{c} \; \forall c \in \mathcal{C}, \; \alpha_{d} \circ \beta_{d} = \beta_{d} \circ \alpha_{d} \; \forall d \in \mathcal{C} \; \Rightarrow \; \alpha \circ \beta = \beta \circ \alpha.$$

Consider the diagram $\require{AMScd}$ \begin{CD} C @>{\alpha_C}>> C \\ @V{\beta_C}VV @VV{\beta_C}V \\ C @>{\alpha_C}>> C. \end{CD} Regard $\beta_C$ as just any morphism. Apply the naturality of $\alpha_C.$