Hahn–Banach extension theorem

First extend by one dimension. Pick $x_0\notin Y$ and let $Y_1=Y\oplus\mathbb{R}x_0$. We must choose the value $c=\Phi(x_0)$ so that $|\varphi(y)+tc|\le M\|y+tx_0\|$ for all $y\in Y,t\in\mathbb{R}$; dividing by $t\ne0$ and substituting $y/t\to y$, this reduces to needing $c$ with $\varphi(y)-M\|y-x_0\|\le c\le M\|y+x_0\|-\varphi(y)$ for all $y\in Y$. Using $\varphi(y_1)-\varphi(y_2)=\varphi(y_1-y_2)\le M\|y_1-y_2\|\le M\|y_1+x_0\|+M\|y_2-x_0\|$, one checks the supremum of the left expressions over $y_1$ never exceeds the infimum of the right expressions over $y_2$, so a valid $c$ in that interval exists; this defines $\Phi$ on $Y_1$ with the same bound $M$.

Next, extend to all of $X$ using Zorn's Lemma. Consider the set of all pairs $(Z,\Psi)$ where $Z$ is a subspace with $Y\subseteq Z\subseteq X$ and $\Psi$ extends $\varphi$ to $Z$ with bound $M$, partially ordered by $(Z_1,\Psi_1)\le(Z_2,\Psi_2)$ iff $Z_1\subseteq Z_2$ and $\Psi_2|_{Z_1}=\Psi_1$. Every chain has an upper bound (take the union of the subspaces and the functional agreeing with each on its domain), so Zorn's Lemma gives a maximal element $(Z^*,\Phi)$.

Finally, if $Z^*\ne X$, the one-dimension extension step above applied to $Z^*$ and any $x_0\in X\setminus Z^*$ would produce a strictly larger admissible pair, contradicting maximality of $(Z^*,\Phi)$. Hence $Z^*=X$, and $\Phi$ is the desired bound-preserving extension to all of $X$.