What's the fastest way to tell if a function is uniformly continuous or not?

Some common situations:

A continuous function $f$ is uniformly continuous if

• $f$ is a map from a closed bounded interval
• if $f$ is a map from any compact set
• If $f$ is differentiable and has a bounded derivative, usually... (see comments on this one. If the domain is the union of finitely intervals that have no common border though, this is true).

A function $f$ is not uniformly continuous if

• $f$ is not continuous
• $f$ has a vertical asymptote (consider $1/x$ on $(0,1)$)
• $f:[a,\infty)\to \mathbb{R}$ is differentiable, and $|f'|$ grows without bound as $x \to \pm \infty$.

Outside of that, I'd say just go back to the definition. I hope you find this helpful.

Probably not as useful as the sufficient conditions listed above, but the necessary and sufficient condition for a function $f$ to be uniformly continuous is that whenever you have two sequences $(x_n)$ and $(y_n)$ such that $d(x_n,y_n)$ goes to $0$ (here $d$ stands for the distance, which is a notion pertaining to the general theory of metric spaces), then the same happens to $d\bigl(f(x_n),f(y_n)\bigr)$. Note that is not required the sequences $(x_n),(y_n)$ to have some "nice" behavior (convergence, Cauchy condition, boundedness, etc.) other than their corresponding terms come closer.

One more condition that may be intuitively/visually compelling: If $U \subset \mathbf{R}^n$ is bounded and $f:U \to \mathbf{R}$ is continuous, then $f$ is uniformly continuous on $U$ iff there exists a continuous extension of $f$ to the closure of $U$.

(Consequently, a bounded, continuous, monotone function on an interval of reals is uniformly continuous (even if the domain is unbounded); $f(x) = \sin(1/x)$ is not uniformly continuous on $(0, 1)$; $\operatorname{sgn}(x) = x/|x|$ is not uniformly continuous on $\mathbf{R}\setminus\{0\}$, but is uniformly continuous on $\mathbf{R}\setminus[0, \delta]$ for every $\delta > 0$, etc.)