Recall that a local ring is a ring with a unique maximal ideal, and that a field is a ring where $0$ is maximal. So a discrete valuation ring is equivalently a local ring which is also a PID, but which is not a field.
Also recall that a local ring can equivalently be defined as a ring where $0 \neq 1$ and for all $x$, either $x$ or $1-x$ is a unit.
Lemma 1: the localisation of any ring at a prime ideal is a local ring.
Proof: Consider a prime ideal $p \subseteq R$. Suppose $0 = 1$ in $R_p$. Then there exists some $x \notin p$ with $0 = x$. But then $0 \notin p$, which contradicts that $p$ is an ideal.
Now consider some $\frac{a}{b} \in R_p$. Then either $a \notin p$ or $b - a \notin p$, since if both were in $p$, we’d have $a + (b - a) = b \in p$. So either $a$ is invertible in $R_p$, in which case $\frac{a}{b}$ is, or $b - a$ is invertible in $R_p$, in which case $1 - \frac{a}{b} = \frac{b - a}{b}$ is. So $R_p$ is indeed a local ring. $\square$
Lemma 2: Let $R$ be a PID. Than any nonzero localisation of $R$ is also a PID.
Proof: any nonzero localisation of an integral domain is an integral domain. Now consider an ideal $I \in S^{-1} R$. Then consider $I \cap R$, which is an ideal of $R$, hence principal. Then write $I \cap R = pR$. I claim $I = (p)$. For clearly $p \in I$. And if $\frac{a}{b} \in I$, then $a \in I \cap R$, so we can write $a = pc$ and thus $\frac{a}{b} = p \frac{c}{b}$. $\square$
Combining these two lemmas, we see that localising on a prime ideal gives us a local ring which is also a PID. However, if our original ring $R$ is a field, then localising the field at a maximal ideal will give us the same field, whose maximal ideal is zero. So we must add a condition to your claim, which is that the prime ideal in question be nonzero. In that case, write the ideal as $(p)$. Then $p$ is not invertible in $R_p$, nor is it zero. So $R_p$ is not a field, and thus its unique maximal ideal is nonzero.