Proving the set of distances is connected

Question

Let A be a connected subset of a metric space X and let $y \in X$. Let $E\subseteq \mathbb{R}$ be defined as $E=\{d(a,y):a\in A\}$. Show that E is connected.

This is what I got so far:

Assume E is not connected, then E is not an interval. Thus $\exists a_1,a_2 \in A$ and $ t\in \mathbb{R}, t\notin A $ such that $d(a_1,y)<t<d(a_2,y)$. I'm not sure how to continue, any suggestions/hints are appreciated.

Answer 1

Let $U=\{x \in A: d(x,y) <t\}$ and $V=\{x \in A: d(x,y) >t\}$. These are disjoint and their union is $A$ because $t$ does not belong to $E$. I leave it yo you to verify that these two sets are open. Since $a_1 \in U$ and $a_2 \in V$ these two sets are nonempty, contradicting the fact that $A$ is connected. Hint for showing that $U$ is open: let $d(x_0,y) <t$. Verify that $B(x_0,r) \subset U$ if $0<r<t-d(x_0,y)$.

Answer 2

$\textbf{Hint}$ What can we say about the image of a connected set under a continuous map?