If whenever $|x-a|<\delta$, $|y-a|<\delta$ then $|f(x)-f(y)|<\epsilon$, is this equiv. to $f$ cont. at $c$ then $f$ cont. on some interval around $c$? Is the following theorem

If $f$ is continuous at $a$, then for any $\epsilon>0$ there is a
$\delta>0$ so that whenever $|x-a|<\delta$ and $|y-a|<\delta$, we have
$|f(x)-f(y)|<\epsilon$.

equivalent to saying that if $f$ is continuous at a point $c$ then it is continuous on some interval around that point $c$?
 A: Take the function
$$
f : x \mapsto \begin{cases}
0 & -\infty <x \le 0 \\
q^n &  q^{n+1} \le x < q^{n} ;& n\in \mathbb N, 0 < q < 1
\end{cases}
$$
which is continuous at $x = 0$, but not on any open intervall containing $x=0$. See this for $q = 1/2$

A: Thomae's function, defined by
$$ f(x) = \begin{cases}
\dfrac{1}{q} & \text{if $x=\dfrac{p}{q}\in\mathbb{Q}$ with $\gcd(p,q) =1$, and} \\[1.5ex]
0 & \text{otherwise},
\end{cases}$$
is an example of a function which acts as a counterexample to the purported equivalence.  This function is continuous at every irrational number, and discontinuous at every rational number.  Hence at every point $a$ such that $f$ is continuous at $a$, there is no interval containing $a$ such that $f$ is continuous on that interval.
The proposition you have written is saying something different.  Essentially, it says that if $f$ is continuous at $a$ and if $x$ and $y$ are both "close" to $a$, then $f(x)$ and $f(y)$ are "close" to each other—a continuous function cannot vary too much on some small interval around $a$.  A proof is as follows:

Proof:  Suppose that $f$ is continuous at $a$ and that $\varepsilon > 0$.  As $f$ is continuous at $a$, there exists some $\delta > 0$ such that
$$ |b-a| < \delta \implies |f(b) - f(a)| < \frac{\varepsilon}{2} $$
for any $b$ in the domain of $f$.  Then if
$$ |x-a| < \delta \qquad\text{and}\qquad |y-a| < \delta, $$
it follows that
\begin{align} |f(x) - f(y)|
&= |f(x) - f(a) + f(a) - f(y)| && \text{(add zero)} \\
&\le |f(x) - f(a)| + |f(y) - f(a)| && \text{(triangle inequality)} \\
&< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\
&= \varepsilon. \end{align}

The proposition can be proved (it is a true statement), while the statement "if $f$ is continuous at $c$, then $f$ is continuous on an interval containing $c$" has counterexamples (it is a false statement).  Therefore the proposition cannot be equivalent to the second statement.
