Does there exist a sequence of sets such that the sequence of their cardinalities is strictly decreasing? Does there exist a sequence of sets such that the sequences of their cardinalities is strictly decreasing?
More explicitly, does there exist a sequence of sets $S_1,S_2...$ such that for each $i$ there exists an injection from $S_{i+1}$ into $S_i$ yet $S_i$ and $S_{i+1}$ is no bijection? 
In other words, each $S_{i+1}$ is strictly smaller (in cardinality) than $S_i$.
 A: No: cardinalities are well-ordered, and no well-order admits an infinite descending chain. Such a chain would be a non-empty set with no least element.
A: Assuming the axiom of choice, which is the usual assumption, as Brian notes, the answer is negative. Cardinals are well-ordered, and that's that.
Without the axiom of choice, however, the situation gets more complicated.
It is possible that there is a sequence of non-empty families, $\cal S_i$, such that any two $S_1,S_2\in\cal S_i$ are of the same cardinality, i.e. there is a bijection between the two sets, and for every $i$, and $S\in\cal S_i$ and $T\in\cal S_{i+1}$ there is an injection from $T$ into $S$, but not in the other direction.
But, and here's the funny part, there is no function choosing from all the $\cal S_i$'s at the same time. That is, we cannot convert this decreasing sequence of cardinals into a decreasing sequence of sets.
Nevertheless, in this situation we can usually find a different sequence of sets which are decreasing. 
Okay, you say, then let's ask this. Suppose that the axiom of choice fails. Can we always find some sequence like this? Well, this is an open question. 
Most, if not all, models we know where the axiom of choice fails will have such sequences, so at the very least, pretty much all the weak versions of the axiom of choice that we know of are not sufficient to prove that there are no such decreasing sequences.
