The empty space is the topological space with no points. That is, it is the empty set equipped with its unique topology.
The empty space is the initial object in Top. It satisfies all separation, compactness, and countability conditions (separability, first countability, second-countability). It is also both discrete and indiscrete, a distinction it shares only with the point.
Debate rages over whether the empty space is connected (and also path-connected). With the common naive definitions that “a space is connected if it cannot be partitioned into two disjoint nonempty open subsets” and “a space is path-connected if any two points in it can be joined by a path,” the empty space is trivially both connected and path-connected.
However, in some ways these definitions are too naive. The question of whether the empty set is (path-)connected is analogous in many ways to the question of whether $1$ is prime. The above definitions are then analogous to saying that “a natural number $p$ is prime if any factor of it is either equal to $1$ or to $p$,” according to which $1$ is prime—but there are better definitions that exclude $1$.
For instance, we may say that “$p$ is prime if it has exactly two factors, itself and $1$;” with this definition $1$ is not prime, since it has exactly one factor. Likewise, we may say that a space is (path-)connected if it has exactly one (path-)component?; with this definition the empty space is not connected, since it has exactly zero components. (Lest you question that last statement, note that the correct definition of a (path-)component of a space $X$ is as an equivalence class of points of $X$ under some equivalence relation. There is a unique equivalence relation on the empty set, and it has zero equivalence classes.)
Here are some other reasons why the empty space should not be considered (path-)connected:
If the empty space were (path-)connected, unique decomposition into (path-)connected components would fail: $X \cup Y = \emptyset \cup X \cup Y = \dots$. This is analogous to how if $1$ were a prime, then unique factorization into primes would fail: $6 = 2 \cdot 3 = 1 \cdot 2 \cdot 3 = 1 \cdot 1 \cdot 2 \cdot 3 = \dots$.
In homotopy theory, one defines a space $X$ to be $k$-connected if $\pi_i(X)$ is trivial (that is, has exactly one element) for $i \le k$. When $k =0$ this says that $\pi_0(X)$ should have exactly one component—that is, that $X$ should be path-connected. (Actually, this definition really only makes sense if we phrase it in terms of homotopy groupoids; homotopy groups are only defined once we choose a basepoint, which is clearly impossible for the empty space.)
Category-theoretically, one may say that a space $X$ is connected if the functor $hom(X,-)$ preserves coproducts. Since $\hom(\emptyset,-)$ is constant at the point, it certainly does not preserve coproducts.
The statement that a product $X\times Y$ is connected if and only if its components $X$ and $Y$ are connected fails if the empty set is regarded as connected.
See too simple to be simple for general theory.