Application of corollary of Banach-Steinhaus theorem Here is the corollary of the Banach-Steinhaus theorem:

Corollary A: Let $A\subseteq X$ be a set in $X$ such that for every $f\in X^*$ there exists a constant $C(f)$ such that $\sup_{x\in A}|f(x)|\leq C(f)$. Then $A$ is norm bounded, i.e., there exists a constant $C$ such that $\|x\|\leq C$ for all $x\in A$.

The author then applies Corollary A via the following theorem and proof:

Theorem 1: Let $H$ be a Hilbert space and let $A\colon H\to H$ be a linear operator with $\text{Dom}A=H$ (not necessarily bounded) and $A=A^*$, i.e.,
$\langle Ax,y \rangle= \langle x,Ay \rangle$. Then $A$ is bounded.
Proof. We have
$$
|\langle Ax,y \rangle| = |\langle x,Ay \rangle|\leq\|x\|\cdot\|Ay\|\leq\|Ay\|
$$
for all $x\in H$ with $\|x\|\leq1$ and for all $y\in H$. By Corollary A we conclude that the set $\{Ax:\|x\|\leq1\}$ is bounded which implies $\|Ax\|\leq C\|x\|$.

However this proof confused me. Can anyone really explain what is going on in this proof, i.e., what is the set $A$ supposed to be from Corollary A. Also, why does this even finish the proof? Don't we have to show the operator $A$ is bounded for every $x\in H$, not just those $x$ with $\|x\|\leq1$? Or does this finish to proof since it's enough to show the operator $A$ is bounded for $x=0$, and clearly $\|x=0\|\leq1$...?

I should add that, I take it the set $A$ is supposed to be $A:=\{Ax:\|x\|\leq1\}$; however, then this set being norm bounded means there exists $C>0$ such that $\|Ax\|\leq C$ for all $Ax\in A$, so also $\|x\|\leq 1$. But how does $\|Ax\|\leq C$ together with $\|x\|\leq1$ give $\|Ax\|\leq C\|x\|$?
 A: It's unfortunate that $A$ is overloaded to be both a set and an operator. Let me call the linear operator $T$ instead.
Define the set $A = \{Tx: \|x\| \leq 1\}$. Since every linear functional on $H$ is of the form $\langle \cdot, y\rangle$ by the Riesz Representation Theorem, the calculation in the proof of Theorem 1 tells you that for every $f \in H^*$, $sup_{\{Tx: \|x\| \leq 1\}} |f(Tx)| \leq C(f) = \|Ty\|$ where $y$ is the element of $H$ corresponding to $f$ via the Riesz Representation Theorem.
Hence, by Corollary A, $\{Tx: \|x\| \leq 1\}$ is a bounded set. Therefore the operator norm of $T$ is finite.

Rereading the question, it seems you might not have seen the various equivalent representations of the operator norm. In this case, for arbitrary $x \neq 0$, let $z = \|x\|^{-1} x$. Then $z$ has norm at most $1$ so that $\|Tz\| \leq C$ for the constant $C$ bounding the set above. Multiplying through by $\|x\|$ yields $\|Tx\| \leq C\|x\|$ as desired.
In general, it might be worth googling/checking a standard text to see the various equivalent forms of the operator norm if you have not seen these before.
