Let $\varnothing\ne M\subset\mathbb R^n$ be a convex compact set, suppose $M\subseteq B_r(u)$ (closed ball).
Then the map\begin{align}f:\ B_r(u)&\longrightarrow M\newline x&\longmapsto f(x) = \arg\min_{y\in M}\|x-y\|\end{align}is continuous.
I was given a proof as followed :
First, the map $f$ is well-difined due to $f(x)$ exists and is unique.
For any $x\in B_r(u)$ and any sequence $x_n\to x$, by definition of $f$ we have\begin{align}\|x_n-f(x_n)\|\leq \|x_n-y\|, \ \forall y\in M;\tag1\end{align}Hence, if $a$ is any limit of $f(x_n)$ then\begin{align}\|x-a\|\leq \|x-y\|, \ \forall y\in M. \tag2\end{align}This proves that $f(x)$ is the only limit point of $\{f(x_n)\}$ which lies in the compact set $M$. Hence $f(x_n)\to f(x)$, this shows the continuity of $f$.
There're some points of the proof that I still yet to comprehend. First, I understand the existance of $f(x)$ due to the continuity of the function $y\longmapsto\|x-y\|$ on the compact $M$. However, I didn't see why $f(x)$ is unique.
Next, I don't understand why from (1) we got (2). In concrete, I don't get why we assumed that $a$ is the limit of $f(x_n)$. What if $f(x_n)$ isn't convergent ?
I really hope someone would help me to explain it more explicitly. Thanks.