Proof: Let $\eps >0$ and $\p$ denote any point. Suppose $\x$ satisfies $\norm{\x-\p} < \eps$.

\[\begin{alignat*}{2} \tag*{$\tcirc{1}$} \norm{\x} &\leq \norm{\x-\p} + \norm{\p} &\hspace{4em}& \href{norm.html}{\text{Triangle inequality}} \\ \tag*{$\tcirc{2}$} \norm{\p} &\leq \norm{\p-\x} + \norm{\x} &\hspace{4em}& \href{norm.html}{\text{Triangle inequality}} \\ \norm{\x} - \norm{\p} &< \eps && \tcirc{1} \\ \norm{\p} - \norm{\x} &< \eps && \tcirc{2}; \href{norm.html}{\text{Homogeneity}} \end{alignat*}\]

I.e., there exists $\delta$ (in this case, $\delta = \eps$) such that $\norm{\x-\p} < \delta \implies \abs{f(\x)-f(\p)} < \eps$.