Is every subset of a lattice a lattice?
Lattice consists of a partially ordered set in which every two elements have to have unique supremum and infimum.
I'm confused about what the answer is. I considered a lattice $(L, \le)$ where $L$ is a set {1, 2, 3, 6} and $\le$ is relation of divisibility (a simplified version of this example) (e.g. 1 divides 2, 3 and 6, 2 divides 6, etc.).
Now if I take $\{1, 2, 3\}$ as a subset of this lattice, would this still be a lattice? Would $2$ and $3$ be missing a supremum? On one hand I think they would because $6$ is no longer in the set but on the other hand isn't supremum not required to be in the set in question? For example the set of real numbers on an open interval $(0, 1)$ has a supremum equal to 1, even though 1 is not in the set.
The example you gave is correct and also the smallest. Your poset is a rombus, where $6$ is the top vertex, $2$ and $3$ the mid ones, and $1$ the bottom vertex.
If you remove either the top vertex or the bottom you lose the existence of supremum or infimum respectively.
This is only a small correction to Ethan Bolker's comment. The open unit interval, as any totally ordered set is a lattice. The definition of lattice only requires the existence of join and meet for pairs of elements (and by induction for finite subsets). In a totally ordered set supremum of two elements is the maximum among them, and the infimum is the minimum among them.
In particular the totally ordered sets are lattices for which every subset is also a lattice.