Is it true that $d((f+g)(x), (f+g)(y)) \leq d(f(x),f(y)) + d(g(x),g(y))$? In arbitrary metric space $(M, d)$, is it true that $d((f+g)(x), (f+g)(y)) \leq d(f(x),f(y)) + d(g(x),g(y))$?
Clearly, in the simple case where $M = \mathbb{R}$ and $d(x,y) = | x - y |$, we have
$$\begin{align}
d((f+g)(x), (f+g)(y)) &= |(f+g)(x) - (f+g)(y)| \\
&= |(f(x) + g(x)) - (f(y) + g(y))| \\
&= |(f(x) - f(y)) + (g(x) - g(y))| \tag{1} \\
&\leq |f(x) - f(y)| + |g(x) - g(y)| \\
&= d(f(x), f(y)) + d(g(x), g(y)),
\end{align}$$
as desired. So basically, this proof requires the ability to rearrange the values within the absolute values sign, as indicated in line (1), which is certainly possible.
But when I try this same type of proof using just the properties of an arbitrary metric, I get
$$\begin{align}
d((f+g)(x), (f+g)(y)) &= d(f(x) + g(x), f(y) + g(y)) \\
&(=?) \ |d(f(x), f(y)) -  d(g(x), g(y))| \tag{2}\\
&\leq d(f(x),f(y)) + d(g(x), g(y)),
\end{align}$$
and I'm not sure I can do the same type of rearrangement in an arbitrary metric space, which is why I've put the question mark in line (2). I feel like I'm missing something obvious! (For what it's worth, I'm trying to use this to prove that the sum of uniformly continuous functions is uniformly continuous.)

Its easy to show that the theorem holds when the metric is induced by a norm(as you showed). Consider the discrete metric $d(x,y)=1 \iff x=y$ and $0$ otherwise on $\mathbb{R}$. Let $f(x)=g(y)=1$, $f(y)=g(x)=2$. Then
$$\text{LHS}=d(1+2,1+2)=1\\
  \text{RHS}=d(1,2)+d(2,1)=0$$
