Do the homomorphisms really have to be continuous? I read that
If $\varphi, \psi$ are continuous homomorphisms from a normed algebra $A$ to a normed algebra $B$ then $\varphi = \psi$ if $\varphi$ and $\psi$ are equal on a set $S$ that generates the algebra $A$. 
Do they really have to be continuous? As far as I can tell even if they are not but agree on $S$ and $S$ generates $A$ then they still have to be equal. The equality seems to follow from the properties of homomorphisms. 
 A: You are right to question this. The issue is likely an ambiguous use of "generates." If $S$ generates $A$ abstractly as an $\mathbf{R}$-algebra or a $\mathbf{C}$-algebra or whatever, then yes, as you surmise, there is no need to assume continuity. But sometimes, in the context of, e.g., Banach algebras or more general topological groups, people use the term "generates" when they really mean "topologically generates." This means that the abstract subalgebra of $A$ generated by $S$ is dense in $A$. So this is generally strictly weaker than $S$ generating $A$ abstractly as an algebra. If $S$ only topologically generates $A$ as an algebra, then to conclude that $\varphi=\psi$ if $\varphi\vert_S=\psi\vert_S$, you indeed need to assume continuity (and one needs also to assume that $B$ is Hausdorff, but since it is normed, it will be).
EDIT (to respond to the comment made by the OP): The set $\{a\in A:\varphi(a)=\psi(a)\}$ is the inverse image of the diagonal $\Delta\subseteq B\times B$ under the ring map $(\varphi,\psi):A\rightarrow B\times B$ given by $a\mapsto(\varphi(a),\psi(a))$. It is therefore a subalgebra of $A$ which, if $B$ is Hausdorff, is moreover closed in $A$. Therefore, if it contains $S$, i.e., if $\varphi\vert_S=\psi\vert_S$, then it contains the closure of the subalgebra generated by $S$, which, by assumption, is all of $A$, so $\varphi=\psi$.
