If $A:=\left\{z \in \mathbb C\ \colon \left|z\right|\leqslant 1\right\}$ and $f\ \colon A\to \mathbb C$ continuous, let $f_r(z)=f(rz)$ for all $z$ in $A$ and $0<r<1$.
Now I have to prove that $f_r\to f$ uniformly on $A$ if $r\to 1$.
I don't know how to start. Would somebody help with this question, please?
Uniform convergence over an interval-indexed collection of functions is the same as uniform convergence over a countable collection. That is, you must prove:
Think about the bit $|f_r(z)-f(z)|$. This is just $|f(rz)-f(z)|$. By continuity of $f$, for fixed $z$, if $|z-w|$ is sufficiently small, say $|z-w|<\delta(\epsilon, z)$, then we can make $|f(z)-f(w)|<\varepsilon$. It's not very hard to make $|z-rz|$ small; you should be able to find $R$ such that $r>R$ implies $|z-rz|<\delta(\varepsilon, z)$.
But this isn't enough, since this $\delta(\varepsilon, z)$ apparently depends on the value of $z$. But in fact $f$ is uniformly continuous. If you can see why, then the remainder of the proof should be clear.