Let $\phi$ be continuous in a neighborhood of $0\in\mathbf{R}^3$ (you may assume it to be uniformly continuous, if you like). Do we have that $$\lim_{\epsilon\rightarrow 0}\frac{1}{4\pi\epsilon^2}\int_{\partial B_{\varepsilon}(0)}\phi \ do=\phi(0)?$$ My "proof" would be to write $$\left|\frac{1}{4\pi\epsilon^2}\int_{\partial B_{\varepsilon}(0)}\phi-\phi(0) \ do\right|\leq\sup_{x\in\partial B_\epsilon(0)}\left|\phi(x)-\phi(0)\right|\stackrel{\epsilon\rightarrow 0}{\longrightarrow}0$$ but does this last limit even hold? Under which circumstances would it be true?
This result is used in proving $1/4\pi r$ to be a fundamental solution of $-\Delta$.
Your proof works well. Under the sole assumption that $\phi$ is integrable over the sphere, we have
$$\frac{1}{4\pi\varepsilon^2}\left\lvert \int_{\partial B_\varepsilon(0)} \phi(x) - \phi(0)\,do \right\rvert \leqslant \frac{1}{4\pi\varepsilon^2}\int_{\partial B_\varepsilon(0)} \lvert \phi(x)-\phi(0)\rvert\,do \leqslant \sup_{x\in\partial B_\varepsilon(0)} \lvert \phi(x) - \phi(0)\rvert.$$
If $\phi$ is continuous in a neighbourhood of $0$, then $\phi$ is integrable over all small enough spheres, and
$$\lim_{\varepsilon\searrow 0} \sup_{x\in\partial B_\varepsilon(0)} \lvert \phi(x) - \phi(0)\rvert = 0$$
is a direct consequence of the continuity in $0$.