Suppose $g: R \to R$ is a continuous function such that $g(0) = 0$, and suppose $f : R\to R$ is such that$|f(x) - f(y)|\leq g( x - y) $ ... Suppose $g: R \to R$ is a continuous function such that $g(0) = 0$, and suppose $f : R\to R$ is such that $|f(x) - f(y)| \leq g( x - y) $ for all $x$ and $y$. Show that $f$ is continuous. 
I really need help... 
this is what I got
$|x - 0  | < \delta / 2 \Rightarrow | g(x) - g(0) | = | g(x) |  < \epsilon /2 $
Similarly $\mid y -0 \mid $... 
and then $ \mid x - y \mid  = \mid x - 0 + 0 - y \mid = \mid x - 0 \mid + \mid y - 0 \mid < \delta$
and then $|g(x-y) - g(0) | = | g(x-y) - 0 | = | g(x) - g( y) | = | g(x) - g(0) + g(0) -g(y) |  \leq | g(x) - g(0) | + | g(x) - g(y) | < \epsilon$
= > $|g(x-y) | < \epsilon$
so I can say that $\mid f(x) - f(y) \mid < \epsilon$
Is this right? 
 A: Use continuity of g at zero to get uniform continuity of f.
A: The content of the givens is the following: If $x$ is near $y$ then $|x-y|$ is small. Therefore $x-y$ is near zero, and so is $g(x-y)$. It follows that $|f(x)-f(y)|\leq g(x-y)$ is small as well. Now we have to build a logic $\epsilon$-$\delta$-chain establishing the continuity of $f$ according to the definition.
Let an $\epsilon>0$ be given. Since $g(0)=0$, and $g$ is continuous at $0$, there is a $\delta>0$ such that $|g(t)|<\epsilon$ as soon as $|t|<\delta$.
We now go back to $f$: Assume that $x$, $y\in{\mathbb R}$ are so near to each other that $|x-y|<\delta$. Then $$|f(x)-f(y)|\leq g(x-y)<\epsilon\ .$$
Since $\epsilon>0$ was arbitrary we have therewith proven that $f$ is actually uniformly continuous on ${\mathbb R}$.
A: Let $\lim_{n \rightarrow \infty}x_n = x$,
$$|f(x_n)-f(x)| \leq g(x_n-x)$$
$$\lim_{n \rightarrow \infty} |f(x_n)-f(x)| \leq \lim_{n \rightarrow \infty} g(x_n-x)=g(\lim_{n \rightarrow \infty}x_n-x)=g(0)=0$$
Hence 
$$\lim_{n \rightarrow \infty} |f(x_n)-f(x)|=0$$
Hence as $\lim_{n \rightarrow \infty}x_n=x \implies \lim_{n \rightarrow \infty} f(x_n)=f(x)$
Edit: 
I just see your attempt, I don't think you can justify that $$|g(x-y)|=|g(x)-g(y)|$$
