[proofplan]
The proof is a direct unpacking of the [operator norm](/page/Operator%20Norm) for the inclusion map. We compare two quantities: the operator norm of $j$ and the set of all constants that bound the $Y$-norm by the $X$-norm. First we show that every admissible constant bounds the operator norm from above. Then we show that the operator norm itself is admissible, so it is the least possible such bound.
[/proofplan]
[step:Define the admissible constants and the operator norm]
Let
\begin{align*}
\mathcal C:=\{C\ge 0:\|x\|_Y\le C\|x\|_X\text{ for every }x\in X\}.
\end{align*}
Since $X\hookrightarrow Y$ is a continuous embedding, the inclusion map $j:X\to Y$ is a bounded [linear map](/page/Linear%20Map), so $j\in \mathcal L(X,Y)$ and $\|j\|_{\mathcal L(X,Y)}<\infty$.
By the definition of the operator norm,
\begin{align*}
\|j\|_{\mathcal L(X,Y)}=\sup\{\|jx\|_Y:x\in X,\ \|x\|_X\le 1\}.
\end{align*}
Because $j$ is the inclusion map, $jx=x$ as an element of $Y$ for every $x\in X$. Hence
\begin{align*}
\|j\|_{\mathcal L(X,Y)}=\sup\{\|x\|_Y:x\in X,\ \|x\|_X\le 1\}.
\end{align*}
[/step]
[step:Show that every admissible constant bounds the operator norm]
Let $C\in\mathcal C$. Then $C\ge 0$, and for every $x\in X$ one has
\begin{align*}
\|x\|_Y\le C\|x\|_X.
\end{align*}
In particular, if $x\in X$ satisfies $\|x\|_X\le 1$, then
\begin{align*}
\|jx\|_Y=\|x\|_Y\le C\|x\|_X\le C.
\end{align*}
Taking the supremum over all $x\in X$ with $\|x\|_X\le 1$ gives
\begin{align*}
\|j\|_{\mathcal L(X,Y)}\le C.
\end{align*}
Since this holds for every $C\in\mathcal C$, the operator norm is a lower bound for $\mathcal C$:
\begin{align*}
\|j\|_{\mathcal L(X,Y)}\le \inf \mathcal C.
\end{align*}
[guided]
We now compare an arbitrary valid embedding constant with the supremum defining the operator norm. Choose any $C\in\mathcal C$. By the definition of $\mathcal C$, this means two things: first $C\ge 0$, and second
\begin{align*}
\|x\|_Y\le C\|x\|_X
\end{align*}
for every $x\in X$.
The operator norm only tests vectors in the closed unit ball of $X$, namely vectors $x\in X$ with $\|x\|_X\le 1$. For such an $x$, the inclusion property gives $jx=x$ as an element of $Y$, and the defining inequality for $C$ gives
\begin{align*}
\|jx\|_Y=\|x\|_Y\le C\|x\|_X\le C.
\end{align*}
Thus $C$ is an upper bound for the set
\begin{align*}
\{\|jx\|_Y:x\in X,\ \|x\|_X\le 1\}.
\end{align*}
Taking the supremum of that set therefore yields
\begin{align*}
\|j\|_{\mathcal L(X,Y)}\le C.
\end{align*}
Because the choice of $C\in\mathcal C$ was arbitrary, every admissible constant lies above $\|j\|_{\mathcal L(X,Y)}$. Hence $\|j\|_{\mathcal L(X,Y)}$ is a lower bound for $\mathcal C$, and therefore
\begin{align*}
\|j\|_{\mathcal L(X,Y)}\le \inf \mathcal C.
\end{align*}
[/guided]
[/step]
[step:Show that the operator norm is itself an admissible constant]
We prove that $\|j\|_{\mathcal L(X,Y)}\in\mathcal C$. Let $x\in X$.
If $x=0$, then
\begin{align*}
\|x\|_Y=0\le \|j\|_{\mathcal L(X,Y)}\|x\|_X.
\end{align*}
If $x\ne 0$, define
\begin{align*}
u:=\frac{x}{\|x\|_X}\in X.
\end{align*}
Then $\|u\|_X=1$, so the definition of $\|j\|_{\mathcal L(X,Y)}$ gives
\begin{align*}
\|ju\|_Y\le \|j\|_{\mathcal L(X,Y)}.
\end{align*}
Using linearity of $j$ and homogeneity of the norm on $Y$,
\begin{align*}
\|x\|_Y=\|jx\|_Y=\left\|j(\|x\|_X u)\right\|_Y=\|x\|_X\|ju\|_Y\le \|j\|_{\mathcal L(X,Y)}\|x\|_X.
\end{align*}
Thus for every $x\in X$,
\begin{align*}
\|x\|_Y\le \|j\|_{\mathcal L(X,Y)}\|x\|_X.
\end{align*}
Since $\|j\|_{\mathcal L(X,Y)}\ge 0$, this proves $\|j\|_{\mathcal L(X,Y)}\in\mathcal C$.
[/step]
[step:Identify the infimum]
Since $\|j\|_{\mathcal L(X,Y)}\in\mathcal C$, one has
\begin{align*}
\inf\mathcal C\le \|j\|_{\mathcal L(X,Y)}.
\end{align*}
The previous lower-bound inequality gave
\begin{align*}
\|j\|_{\mathcal L(X,Y)}\le \inf\mathcal C.
\end{align*}
Combining the two inequalities,
\begin{align*}
\|j\|_{\mathcal L(X,Y)}=\inf\mathcal C.
\end{align*}
Substituting the definition of $\mathcal C$ gives
\begin{align*}
\|j\|_{\mathcal L(X,Y)}=\inf\{C\ge 0:\|x\|_Y\le C\|x\|_X\text{ for every }x\in X\}.
\end{align*}
This is the desired formula.
[/step]
