$f:\mathbb{R}\to\mathbb{R}$ is defined as $f(x)=n\forall x=n\in\mathbb{N}$ Let $f:\mathbb{R}\to\mathbb{R}$ be defined as $ f(x) := \begin{cases} x, & \text{if}\ x\in \mathbb N,\\\\ 0, & \text{else,}   \end{cases} $
and $T=\mathbb{N}\cup\{n+1/n:n\in\mathbb{N}\}$. 
The function $f$ is continuous on $\mathbb{N}$ with respect to the usual metric on the reals, as any function is continuous as every point is an isolated point.  And $f$ is discontinuous on $T$ as $|f(n)-f(n+1/n)|=n>\epsilon=1/2$ say. Am I right?

Actually, your proof is correct but not written down properly.
Let $\epsilon = \frac{1}{2}$ then we claim there is no $\delta > 0$ with
$$|f(x) - f(y)| < \epsilon \qquad \forall\ |x-y| < \delta$$
Let $\delta_0$ be such a choice and chose $N := \lceil \frac{1}{\delta_0} \rceil + 1$, then
$$\left|N - N+\frac{1}{N}\right| = \frac{1}{N} < \delta_0$$
But
$$\left|f(N) - f\left(N+\frac{1}{N}\right)\right| = N > \epsilon$$
So $f$ is not uniformly continuous on $T$, but locally continuous.  
$f$ is (uniformly) continuous on $\mathbb N$, locally cont. on $T$ and discontinuous on $\mathbb R$ in every $x\in\mathbb N$
