Let $(X,d)$ be a metric space and $A,B\subseteq X$ subsets, $x\in X$. Then $$ d(A,B) \le d(x,A) + d(x,B). $$
I have an attempt below but just don't feel it's correct and want some feedback.
Let $t\in A, s\in B$. Then $$ d(t,s) \le d(t,x) + d(s,x) \\ \therefore d(t,s) - d(t,x) \le d(s,x) $$ and $$ d(A,B) - d(t,x) \le d(t,s) - d(t,x) \\ \therefore d(A,B) - d(t,x) \le d(s,x). $$ Once the last inequality stands for all $s\in B$, we have $$ d(A,B) - d(t,x) \le \inf\{d(s,x)|s\in B\} = d(x,B), $$ then $$ d(A,B) - d(x,B) \le d(t,x) $$ for all $t\in A$. Accordingly, $$ d(A,B) - d(x,B) \le \inf\{d(t,x)|t\in A\} = d(x,A), $$ and it follows.