There is a set of continuous functions $f$ on [0, 1] with supremum metric (metric space). Proof that $\phi(f) = f(0) + f(1)$ is continous
Solution 1:
This is just like showing that any other function is continuous.
Fix $f\in C([0,1],\mathbb{R})$. Let $\varepsilon>0$ be given.
We need to find $\delta>0$ such that for $g\in C([0,1],\mathbb{R})$ with $d_{\sup}(f,g)<\delta$, we have $$|\phi(f)-\phi(g)|=|f(1)+f(0)-(g(1)+g(0))|\leq |f(1)-g(1)|+|f(0)-g(0)|<\varepsilon$$
Hopefully this puts you on the right track. I will leave it to you to choose the correct $\delta$.