I assume you're talking about open, closed, and compact subsets of a metric space. If you're just talking about the real line $\mathbb{R}$, then ignore the commentary about abstract metric spaces and just replace the term "neighborhood" with "open interval."
In metric spaces, a set $O$ is open if every point in $O$ is an interior point of $O$. In other words, every point of $O$ can be surrounded by an open ball (neighborhood) which itself lies in $O$. So, for instance, the open intervals $(a,b)$ are open subsets of $\mathbb{R}$, and discs $\{x^2 + y^2 \lt R^2 \}$ are open subsets of $\mathbb{R}^2$.
However, one thing to keep in mind is that in more abstract metric spaces, the open sets can be a little strange-looking. For instance, there are metric spaces where every subset is an open set!
Now, a set $F$ is closed if every accumulation point of $F$ is in $F$. Accumulation points are points $x$ such that every open neighborhood around $x$ must intersect the set in question. So, for instance, the accumulation points of the open disc $\{x^2 + y^2 \lt R^2\}$ are precisely the set of points in the closed disc $\{x^2 + y^2 \leq R^2\}$.
One way to think about closed sets is that they are closed under the operation of taking limits. That is, if one were to take a sequence of points in a closed set, then their limit would still have to be in the set. (This can be made rigorous and should be proven at some point.)
As with open sets, the closed sets of a metric space can still be rather strange-looking, and again, there are examples of spaces in which every subset is closed. Nevertheless, it is always true that finite subsets of metric spaces are closed.
You should note also that if $(X,d)$ is a metric space, then the sets $\emptyset$ and $X$ are always going to be both open and closed. In general, the subsets of a metric space may be open, closed, both or neither. (So just because a set is not open does not make it closed!)
Compact sets are a little less intuitive. The rigorous definition is that a set $K$ is compact if every open cover of $K$ has a finite subcover. Since I don't want to babble for too long, I won't go into what this means in detail. However, I've always intuited compact sets as being "super closed" sets -- they're like closed sets, but even stronger.
It is a fact (whose proof you should learn or create) that every compact subset of a metric space is closed and bounded, but the converse is not necessarily true. In the euclidean spaces $\mathbb{R}^n$, the converse is, in fact, true, but this is not obvious, and is known as the Heine-Borel Theorem.
Unlike open and closed sets, in a metric space $(X,d)$, the set $X$ may not be compact. Therefore, it makes sense to talk about "compact metric spaces," whereas it doesn't make sense to talk about "open" or "closed" metric spaces (since they're all open and closed subsets of themselves).
Uniform continuity is a stronger condition than continuity. That is, every uniformly continuous function is continuous, but not necessarily conversely. It is a theorem (Heine-Cantor) that on compact sets, the converse will hold. So on compact sets, continuity is equivalent to uniform continuity. Again, this is an important theorem whose proof you should learn.
Essentially, uniform continuity of a function $f$ on a set $E$ can be thought of as saying that the "degree of continuity" is the same for every point $x$ in the set $E$. This can be made slightly more precise by saying that the $\delta > 0$ that one gets in the $\epsilon-\delta$ definition of continuity can be chosen to be the same for all $x$ in $E$ -- that is, it will depend only on $\epsilon$, and not on the point $x$. For plain old continuous functions, the $\delta$ will depend on both $\epsilon$ and $x$.