I am trying to show the above statement:
Show that the set of points that are nearer $a$ than $b$ in the sense of Euclidean $\lVert\cdot\rVert_2$ are convex.
My attempt
This follows from the Proposition: The convex hull of $M \subset \mathbf R^2$ is convex, indeed the smallest convex set that contains $M$. Moreover, co($M$) is the set of all convex combinations of points of $M$,
\begin{equation} \text{co}(M) := \{x \in \mathbf R^2 | \exists x_1, x_2, ... , x_N \in M \text{ such that } x = \sum\limits_{i=1}^{N} \lambda_i x_i,\\ \text{ for some } \lambda_1, \lambda_2, ..., \lambda_N, \text{ where } \lambda_1 \geq 0 \forall i, \sum\limits_{i=1}^{N} \lambda_i = 1 \}. \end{equation}
The main idea is based on the idea about the corollary 1.14 in http://www.math.udel.edu/~angell/ch1.pdf:
The convex hull of a compact set in $\mathbf R^n$ is compact.
Assume $C \subset \mathbf R^n$ be compact. See the original source for the skipped steps. The sequence $\{A\}_{k=1}^{\infty}$ has a subsequence $\{AA\}_{j=1}^{\infty}$ which converges to a point of co $(C)$ which shows that this latter set is compact.
If $C$ is closed and convex, it has a smallest/nearest element in a certain sense. It is a simple result for analysis that involves the fact that the function $x \to \lVert x\rVert$ is a continuous map from $\mathbf R^n \to \mathbf R$ and the fact that Cauchy sequences in $\mathbf R^n$ converges.
We need also the theorem: Every closed convex subset of $\mathbf R^n$ has a unique element of minimum norm. See the proof in the source above.
There must be also a simpler way to prove the statement. Is there any sense in my attempt? How can you show the statement?
Let $\phi(x) = \|x-a\|^2-\|x-b\|^2= \|a\|^2-\|b\|^2+2 \langle b-a, x \rangle$. $\phi$ is affine (that is, linear plus a constant) hence convex, and so $\{ x | \phi(x) < 0\}$ is convex.
Note: Since it also follows from the above that $\{ x | \phi(x) > 0\}$ is convex, we see that the sets are open half-spaces separated by the hyperplane $\{ x | \phi(x) = 0\}$.