I recently came across this definition: Let $(X,d)$ be a metric space and $A$ be a nonempty subset of $X$. For each $x\in X$ we define a distance from $x$ to $A$ by the equation
$d(x,A)=\inf\{d(x,a) | a\in A\}$
I would guess that such a function will be continuous. But is there an easy and rigorous way to prove it?
Hint
$d(x,A)=\inf\{d(x,a):a\in A\}\le \inf\{d(x,y):y\in A\}+\inf\{d(y,a):a\in A\}$
so $d(x,A)=\inf\{d(x,a):a\in A\}\le \inf\{d(x,y):y\in A\}+ d(y,A)$
so $d(x,A)- d(y,A)\le \inf\{d(x,y):y\in A\}<d(x,y)$
May be this solution is not easy, but you can prove that the distance is indeed Lispchitz.
Let $\varepsilon >0$, and let $x,y \in X$. By the infimum definition, we have that there exists $z_x,z_y\in A$ such that $$ d(z_x,x)\leq d(x,A)+\varepsilon\ \ \wedge\ \ d(z_y,y)\leq d(y,A)+\varepsilon $$ then \begin{align*} d(x,A) &\leq d(x,z_y)\\ &\leq d(x,y)+d(y,z_y)\\ &\leq d(x,y) +d(y,A)+\varepsilon \end{align*} So, we have that: $$ d(x,A)-d(y,A)\leq d(x,y)+\varepsilon $$ changing the roles of $x$ and $y$, we get: $$ d(y,A)-d(x,A)\leq d(x,y)+\varepsilon $$ then $$ |d(x,A)-d(y,A)|\leq d(x,y)+\varepsilon $$ And taking $\varepsilon \to 0$, we have that $d(\cdot,A)$ is Lipschitz, then is continuous.
Let $U, V$ be subsets of a metric space $M$. Let $\{x_i\}$ be a sequence in $U$ and $\{y_i\}$ be a sequence in $V$.
For a function to be continuous, whenever $D(x_i, x_0) < \delta$ for any $\delta > 0$ and $N_\delta(x_0) \subset U$ we have an appropriate $D(f(x_i), y_0) < \epsilon$ for any $\epsilon > 0$ and $f(N_\delta(x)) \subset N_\epsilon(y_0) \subset V$.
We can also define uniform continuity by proving $f$ to be continuous not just at $x_0, y_0$, but any point in the set.
Can you prove continuity using the epsilon-delta definition?
