Distance to a closed set is continuous.

From $d(x',y)\leq d(x',x)+d(x,y)$, by taking infimum we get $d(x',F)\leq d(x',x)+d(x,F)$. Similarly, $d(x,F)\leq d(x',x)+d(x',F)$. Consequently, $|d(x',F)-d(x,F)|<d(x',x)$.

Here is a more worked example without "taking infimum" using the inverse triangle inequality:

We wish to show for all $x, y \in M$, $\lvert f_F(x) - f_X(y)\rvert \le d(x, y)$.

Case: $d(x, z) \ge d(y, z)$.

Fix the $z$ such that both bounds are true. By the definitions of infimum, we have two bounds (let $\epsilon > 0$):

\begin{align} \forall z \in F \quad &f_F(x) \le d(x, z) \\ \exists z \in F \quad &f_F(x) + \epsilon > d(x, z) \end{align}

By the inverse triangle inequality,

\begin{align} & d(x, z) - d(y,z) \le d(x,y) \\ \implies & f_F(x) - d(y, z) \le d(x,y) \\ \implies & f_F(x) - f_F(y) - \epsilon \le d(x,y) \\ \implies & f_F(x) - f_F(y) \le d(x,y) + \epsilon \end{align}

Since this holds for all $\epsilon > 0$, $f_F(x) - f_F(y) \le d(x,y)$. The case of $d(x,z) < d(y,z)$ is similar, so we have proved $\lvert f_F(x) - f_F(y)\rvert \le d(x, y)$. Now simply apply our definition of continuity using $\delta = \epsilon$ (actually $f_F$ is Lipschitz continuous with constant $1$).