[proofplan]
We write the local ring as the localization of the coordinate ring at the multiplicative set of functions nonvanishing at $p$. The key point is that a fraction $a/b$ is invertible exactly when its numerator also does not vanish at $p$. Thus the non-units are precisely the displayed fractions with numerator vanishing at $p$, and a direct calculation shows that these fractions form an ideal. Since every element outside this ideal is a unit, the ideal is the unique maximal ideal.
[/proofplan]
[step:Localize the coordinate ring at functions nonvanishing at $p$]
Let
\begin{align*}
A:=k[X]
\end{align*}
denote the coordinate ring of $X$. Define
\begin{align*}
S_p:=\{f\in A:f(p)\ne 0\}.
\end{align*}
The set $S_p$ is multiplicatively closed because $1(p)=1\ne 0$ and, for $f,g\in S_p$,
\begin{align*}
(fg)(p)=f(p)g(p)\ne 0.
\end{align*}
By the definition of the local ring of an affine variety at $p$,
\begin{align*}
\mathcal O_{X,p}=S_p^{-1}A.
\end{align*}
Thus every element of $\mathcal O_{X,p}$ is represented by a fraction $a/b$ with $a\in A$ and $b\in S_p$.
[/step]
[step:Check that vanishing of the numerator is independent of the representative]
Suppose
\begin{align*}
\frac{a}{b}=\frac{c}{d}
\end{align*}
in $S_p^{-1}A$, where $a,c\in A$ and $b,d\in S_p$. By equality in a localization, there exists $s\in S_p$ such that
\begin{align*}
s(ad-bc)=0
\end{align*}
in $A$. Evaluating at $p$ gives
\begin{align*}
s(p)(a(p)d(p)-b(p)c(p))=0.
\end{align*}
Since $s(p)$, $b(p)$, and $d(p)$ are nonzero elements of the field $k$, this identity implies
\begin{align*}
a(p)=0 \iff c(p)=0.
\end{align*}
Therefore the condition defining $\mathfrak m_p$ does not depend on the chosen representative of the fraction.
[guided]
We must first check that the displayed set is actually well-defined, because an element of a localization can have many fraction representatives. Suppose
\begin{align*}
\frac{a}{b}=\frac{c}{d}
\end{align*}
in $S_p^{-1}A$, with $a,c\in A$ and $b,d\in S_p$. The definition of equality in a localization says that there is some $s\in S_p$ such that
\begin{align*}
s(ad-bc)=0
\end{align*}
in $A$.
Now evaluate this regular function at the point $p$. Since evaluation at $p$ is a $k$-algebra homomorphism $A\to k$, we get
\begin{align*}
s(p)(a(p)d(p)-b(p)c(p))=0.
\end{align*}
The elements $s(p)$, $b(p)$, and $d(p)$ are nonzero because $s,b,d\in S_p$. Since $k$ is a field, they are invertible in $k$. Hence
\begin{align*}
a(p)d(p)=b(p)c(p).
\end{align*}
Because $b(p)$ and $d(p)$ are nonzero, this equality shows that $a(p)=0$ exactly when $c(p)=0$. Thus whether the numerator vanishes at $p$ is an intrinsic property of the element of $\mathcal O_{X,p}$, not an artifact of the chosen fraction.
[/guided]
[/step]
[step:Characterize the units by evaluating the numerator at $p$]
Let $a/b\in S_p^{-1}A$ with $a\in A$ and $b\in S_p$.
If $a(p)\ne 0$, then $a\in S_p$, so $b/a\in S_p^{-1}A$ is defined and
\begin{align*}
\frac{a}{b}\frac{b}{a}=1.
\end{align*}
Thus
\begin{align*}
\frac{a}{b}
\end{align*}
is a unit.
Conversely, suppose $a(p)=0$ and
\begin{align*}
\frac{a}{b}
\end{align*}
is a unit. Then there exists an element
\begin{align*}
\frac{c}{d}\in S_p^{-1}A,
\end{align*}
with $c\in A$ and $d\in S_p$, such that
\begin{align*}
\frac{a}{b}\frac{c}{d}=1.
\end{align*}
Equivalently,
\begin{align*}
\frac{ac}{bd}=\frac{1}{1}.
\end{align*}
By equality in the localization, there exists $s\in S_p$ such that
\begin{align*}
s(ac-bd)=0.
\end{align*}
Evaluating at $p$ gives
\begin{align*}
s(p)(a(p)c(p)-b(p)d(p))=0.
\end{align*}
Since $a(p)=0$, this becomes
\begin{align*}
-s(p)b(p)d(p)=0.
\end{align*}
But $s(p)$, $b(p)$, and $d(p)$ are nonzero elements of the field $k$, so their product is nonzero, a contradiction. Therefore
\begin{align*}
\frac{a}{b}
\end{align*}
is not a unit when $a(p)=0$.
Hence
\begin{align*}
\left(S_p^{-1}A\right)^\times=\left\{\frac{a}{b}\in S_p^{-1}A:a(p)\ne 0\right\},
\end{align*}
and the non-units are exactly
\begin{align*}
\mathfrak m_p=\left\{\frac{a}{b}\in S_p^{-1}A:a(p)=0\right\}.
\end{align*}
[/step]
[step:Show that the non-units form the displayed ideal]
We verify directly that $\mathfrak m_p$ is an ideal of $S_p^{-1}A$.
First, $0=0/1$ belongs to $\mathfrak m_p$ because $0(p)=0$. If
\begin{align*}
\frac{a}{b}, \frac{c}{d}\in \mathfrak m_p
\end{align*}
then $a(p)=0$ and $c(p)=0$, while $b,d\in S_p$. Their sum is
\begin{align*}
\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}.
\end{align*}
The denominator $bd$ lies in $S_p$, and the numerator satisfies
\begin{align*}
(ad+cb)(p)=a(p)d(p)+c(p)b(p)=0.
\end{align*}
Thus
\begin{align*}
\frac{a}{b}+\frac{c}{d}\in \mathfrak m_p.
\end{align*}
Now let
\begin{align*}
\frac{r}{e}
\end{align*}
be arbitrary, with $r\in A$ and $e\in S_p$, and let
\begin{align*}
v=\frac{a}{b}\in\mathfrak m_p.
\end{align*}
Then
\begin{align*}
\frac{r}{e}v=\frac{ra}{eb}.
\end{align*}
The denominator $eb$ lies in $S_p$, and the numerator satisfies
\begin{align*}
(ra)(p)=r(p)a(p)=0.
\end{align*}
Therefore
\begin{align*}
\frac{r}{e}v\in\mathfrak m_p.
\end{align*}
Taking $r=-1$ and $e=1$ shows that additive inverses of elements of $\mathfrak m_p$ also lie in $\mathfrak m_p$. Since the ring is commutative, this proves that $\mathfrak m_p$ is an ideal.
[/step]
[step:Conclude that $\mathfrak m_p$ is the unique maximal ideal]
The element
\begin{align*}
1=\frac{1}{1}
\end{align*}
does not belong to $\mathfrak m_p$, because $1(p)=1\ne 0$. Hence $\mathfrak m_p$ is a proper ideal.
Let $J$ be any proper ideal of $S_p^{-1}A$. If $J$ contained an element outside $\mathfrak m_p$, then by the unit characterization above $J$ would contain a unit. Any ideal containing a unit is the whole ring, contradicting properness. Therefore every proper ideal of $S_p^{-1}A$ is contained in $\mathfrak m_p$.
It follows that $\mathfrak m_p$ is maximal, and that no other maximal ideal can exist. Thus $\mathcal O_{X,p}=S_p^{-1}A$ is a local ring with unique maximal ideal $\mathfrak m_p$.
[/step]
