[proofplan]
We compare the boundedness of $T$ with the size of its image on the open unit ball. If $T$ is bounded, the [operator norm](/page/Operator%20Norm) inequality immediately bounds every vector $Tx$ with $\|x\|_X<1$. Conversely, if the image of the open unit ball is bounded, we rescale each nonzero vector into the open unit ball and then let the scaling parameter decrease to the vector norm. The same rescaling argument shows that taking the supremum over the open unit ball gives the usual operator norm.
[/proofplan]
[step:Show that a bounded operator sends the open unit ball to a bounded set]
Assume first that $T\in\mathcal L(X,Y)$. Define the set $S\subset Y$ by
\begin{align*}
S:=T(B_X(0,1))=\{Tx:x\in X,\ \|x\|_X<1\}.
\end{align*}
For every $y\in S$, there exists $x\in X$ with $\|x\|_X<1$ and $y=Tx$. By the defining operator norm inequality for bounded linear maps,
\begin{align*}
\|y\|_Y=\|Tx\|_Y\le \|T\|_{\mathcal L(X,Y)}\|x\|_X.
\end{align*}
Since $\|x\|_X<1$, this gives
\begin{align*}
\|y\|_Y\le \|T\|_{\mathcal L(X,Y)}.
\end{align*}
Thus $S$ is bounded in $Y$.
[/step]
[step:Use boundedness on the open unit ball to bound the operator globally]
Assume conversely that $S:=T(B_X(0,1))$ is bounded in $Y$. Then there exists a constant $M\ge 0$ such that
\begin{align*}
\|y\|_Y\le M
\end{align*}
for every $y\in S$.
Let $x\in X$. If $x=0$, then linearity gives $Tx=0$, hence
\begin{align*}
\|Tx\|_Y=0\le M\|x\|_X.
\end{align*}
Assume now that $x\ne 0$. For each $\varepsilon>0$, define
\begin{align*}
z_\varepsilon:=\frac{x}{\|x\|_X+\varepsilon}\in X.
\end{align*}
Then
\begin{align*}
\|z_\varepsilon\|_X=\frac{\|x\|_X}{\|x\|_X+\varepsilon}<1,
\end{align*}
so $Tz_\varepsilon\in S$. Therefore
\begin{align*}
\|Tz_\varepsilon\|_Y\le M.
\end{align*}
Using linearity of $T$ and homogeneity of the norm on $Y$,
\begin{align*}
\|Tx\|_Y=(\|x\|_X+\varepsilon)\|Tz_\varepsilon\|_Y\le M(\|x\|_X+\varepsilon).
\end{align*}
Since this holds for every $\varepsilon>0$, letting $\varepsilon\downarrow 0$ gives
\begin{align*}
\|Tx\|_Y\le M\|x\|_X.
\end{align*}
Thus $T$ is bounded.
[guided]
The only subtle point is that $B_X(0,1)$ is the open unit ball, so a vector $x$ with $\|x\|_X=1$ need not itself lie in the ball. To use the hypothesis for an arbitrary nonzero vector $x\in X$, we shrink it slightly.
Since $S=T(B_X(0,1))$ is bounded, there is a constant $M\ge 0$ such that every $y\in S$ satisfies
\begin{align*}
\|y\|_Y\le M.
\end{align*}
Fix $x\in X$. If $x=0$, then $Tx=0$ by linearity, and the desired estimate is immediate:
\begin{align*}
\|Tx\|_Y=0\le M\|x\|_X.
\end{align*}
Now suppose $x\ne 0$. For each $\varepsilon>0$, define
\begin{align*}
z_\varepsilon:=\frac{x}{\|x\|_X+\varepsilon}.
\end{align*}
This is the scaled version of $x$ that lies strictly inside the unit ball, because
\begin{align*}
\|z_\varepsilon\|_X=\frac{\|x\|_X}{\|x\|_X+\varepsilon}<1.
\end{align*}
Therefore $Tz_\varepsilon\in S$, and the boundedness of $S$ gives
\begin{align*}
\|Tz_\varepsilon\|_Y\le M.
\end{align*}
By linearity,
\begin{align*}
Tx=(\|x\|_X+\varepsilon)Tz_\varepsilon.
\end{align*}
Taking the norm and using homogeneity in $Y$ gives
\begin{align*}
\|Tx\|_Y=(\|x\|_X+\varepsilon)\|Tz_\varepsilon\|_Y\le M(\|x\|_X+\varepsilon).
\end{align*}
The parameter $\varepsilon>0$ was arbitrary, so passing to the limit $\varepsilon\downarrow 0$ yields
\begin{align*}
\|Tx\|_Y\le M\|x\|_X.
\end{align*}
This is exactly the boundedness estimate for the [linear map](/page/Linear%20Map) $T:X\to Y$.
[/guided]
[/step]
[step:Identify the supremum over the open unit ball with the operator norm]
Assume now that $T$ is bounded. Define
\begin{align*}
A:=\sup\{\|y\|_Y:y\in T(B_X(0,1))\}.
\end{align*}
The previous first step shows that $A<\infty$.
For every $x\in B_X(0,1)$, the defining operator norm inequality gives
\begin{align*}
\|Tx\|_Y\le \|T\|_{\mathcal L(X,Y)}\|x\|_X\le \|T\|_{\mathcal L(X,Y)}.
\end{align*}
Taking the supremum over all $x\in B_X(0,1)$ gives
\begin{align*}
A\le \|T\|_{\mathcal L(X,Y)}.
\end{align*}
For the reverse inequality, let $x\in X$ satisfy $\|x\|_X\le 1$, and let $c>1$. Define
\begin{align*}
w_c:=\frac{x}{c}\in X.
\end{align*}
Then
\begin{align*}
\|w_c\|_X=\frac{\|x\|_X}{c}<1,
\end{align*}
so $Tw_c\in T(B_X(0,1))$. Hence
\begin{align*}
\|Tw_c\|_Y\le A.
\end{align*}
By linearity and homogeneity,
\begin{align*}
\|Tx\|_Y=c\|Tw_c\|_Y\le cA.
\end{align*}
Letting $c\downarrow 1$ gives
\begin{align*}
\|Tx\|_Y\le A.
\end{align*}
Taking the supremum over all $x\in X$ with $\|x\|_X\le 1$ and using the definition of the operator norm yields
\begin{align*}
\|T\|_{\mathcal L(X,Y)}\le A.
\end{align*}
Combining the two inequalities proves
\begin{align*}
\|T\|_{\mathcal L(X,Y)}=A=\sup\{\|y\|_Y:y\in T(B_X(0,1))\}.
\end{align*}
This completes the proof.
[/step]
