Show that the sum of two uniformly continuous functions is uniformly continuous in an arbitrary metric space

Question

I've seen the proof when you're talking about $\mathbb{R}$ and the metric is the absolute value function, but I'm trying to prove it for an arbitrary metric space. I figure the proof is pretty similar, for a given $\epsilon$, you choose $\delta$ to be the minimum of the $\delta$s required to make $d(f(x), f(y)) < \frac{\epsilon}{2}$ and $d(g(x), g(y) < \frac{\epsilon}{2}$ when $d(x, y) < \delta$. But then I don't know how to argue that $d(f+g(x), f+g(y)) \leq d(f(x), f(y)) + d(g(x), g(y))$. Any help?

Answer 1

EDITED:

After the comments of @Brian Moehring,

We assume that the distance is stable under translation and that both $f$ and $g$ target the same space,

$d(f(x)+a, f(y)+a)=d(f(x), f(y))$, So for every $x,y$ such that $d(x, y) < \delta$ we have,

$d(f(x)+g(x), f(y)+g(y)) \leq d(f(x)+g(x), f(y)+g(x))+d(f(y)+g(x), f(y)+g(y))\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$