It should be straight forward to show that there are infinitely many NP-hard problems:
Proof: Take the problem Remove 1 Vertex 3-COL ($R1V3COL$) which takes a graph $G=(V,E)$ as an instance and yields a yes answer iff a vertex $x \in V$ exists which, when removed from $V$ yields a new graph $G'=(V\backslash\{x\}, \{(u,v) \in E\,|\, u \neq x \land v\neq x\}$ which is a positive instance of $3COL$.
$R1V3COL$ can indeed be reduced to 3-colorability problem (which is proven to be NP-complete) by (as an example reduction) simply removing a vertex from $G$ and testing if $G$ is 3-colorable. If $G$ is not 3-colorable remove another vertex from $G$ and test it. Repeat until there are no vertices left for testing.
Therefore we know $R1V3COL$ is an NP-hard problem.
We can now reduce $R2V3COL$ to a $R1V3COL$ problem (by a similar concept as shown above for $R1V3COL$ to $3COL$ to show that $R2V3COL$ is in NP-hard and so on for every Remove n Vertices 3-COL problem. In other words we can always reduce $R(n)V3COL$ to $R(n-1)V3COL$. Therefore we know that there have to be infinitely many NP-hard problems and we are done.
Now to my Lemma: I cannot think of a simple proof to show that a certain problem with infinitely many variations (like $R(n)V3COL$) is also in NP to prove NP-completeness and therefore prove that infinitely many problems are in the subclass NP-complete.