The Dirichlet function is continuous at no point of $\mathbb{R}$, and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t(c)$

Claim 1. Let $c$ be real and let $\delta >0$ be real. The neighbourhood $N_{\delta }(c)$ contains a rational $u$ and an irrational $v$, and ${\mathbf{1}}_{\mathbb{Q}}(u)-{\mathbf{1}}_{\mathbb{Q}}(v)=1-0=1$; since ${\mathbf{1}}_{\mathbb{Q}}(c)$ is $0$ or $1$, one of $\mathrm{\mid }{\mathbf{1}}_{\mathbb{Q}}(u)-{\mathbf{1}}_{\mathbb{Q}}(c)\mathrm{\mid }$ and $\mathrm{\mid }{\mathbf{1}}_{\mathbb{Q}}(v)-{\mathbf{1}}_{\mathbb{Q}}(c)\mathrm{\mid }$ equals $1$. So the continuity condition at $c$ fails for $\epsilon =1$, no $\delta$ witnessing it, and ${\mathbf{1}}_{\mathbb{Q}}$ is continuous at no point.

A separation estimate. For a real $c$ and a natural $q\ge 1$ put $m:=\lfloor qc\rfloor$ and define $d_q(c):=1\mathrm{/}q$ if $qc=m$, and $d_q(c):=\min \{qc-m,m+1-qc\}\mathrm{/}q$ otherwise. In both cases $d_q(c)>0$.

A lower bound. For every real $\delta >0$ the neighbourhood $N_{\delta }(c)$ contains an irrational $v$, and $c\in N_{\delta }(c)$, so ${\omega }_t(N_{\delta }(c))\ge \mathrm{\mid }t(c)-t(v)\mathrm{\mid }=t(c)$; taking the infimum over $\delta$ gives ${\omega }_t(c)\ge t(c)$.

With $d_q(c)$ as in step 1.2: for every integer $p$ with $p\mathrm{/}q\mathrm{\ne }c$ one has $\mathrm{\mid }c-p\mathrm{/}q\mathrm{\mid }\ge d_q(c)$. If $qc=m$ then $p\mathrm{\ne }m$, so $\mathrm{\mid }qc-p\mathrm{\mid }=\mathrm{\mid }m-p\mathrm{\mid }\ge 1$; otherwise $m<qc<m+1$, and $p\le m$ gives $qc-p\ge qc-m$ while $p\ge m+1$ gives $p-qc\ge m+1-qc$. Dividing by $q>0$ gives the claim.

For a real $c$ and a natural $N\ge 1$ put ${\delta }_N(c):=\min \{d_1(c),\dots ,d_N(c)\}$, the minimum of a nonempty finite set of positive reals, so ${\delta }_N(c)>0$.

If $x$ is rational with $0<\mathrm{\mid }x-c\mathrm{\mid }<{\delta }_N(c)$ then $q(x)>N$ and hence $t(x)<1\mathrm{/}N$. Indeed $x=p\mathrm{/}q$ with $q:=q(x)$ and $p:=qx$; if $q\le N$ then $p\mathrm{/}q=x\mathrm{\ne }c$, so step 2.1 gives $\mathrm{\mid }c-x\mathrm{\mid }\ge d_q(c)\ge {\delta }_N(c)$, contrary to the hypothesis. So $q(x)>N$, and $t(x)=1\mathrm{/}q(x)<1\mathrm{/}N$ because the canonical inclusion preserves strict order.

An upper bound for $t$ near $c$. Let $\epsilon >0$ be real, take $N\ge 1$ with $1\mathrm{/}N<\epsilon$ and put $\delta :={\delta }_N(c)$. Every $x\in N_{\delta }(c)$ satisfies $0\le t(x)\le M$ where $M:=\max \{t(c),\epsilon \}$: for $x=c$ this is $t(c)\le M$; for $x\mathrm{\ne }c$ rational it is $t(x)<1\mathrm{/}N<\epsilon \le M$ by step 3.1; and for $x$ irrational it is $t(x)=0$.

Hence ${\omega }_t(N_{\delta }(c))\le M$ with $\delta$ and $M$ as in step 4.1, since $\mathrm{\mid }t(x)-t(y)\mathrm{\mid }\le M$ for all $x,y\in N_{\delta }(c)$, so $M$ is an upper bound of the set whose supremum ${\omega }_t(N_{\delta }(c))$ is; and therefore ${\omega }_t(c)\le M=\max \{t(c),\epsilon \}$.

Claim 2 now follows in the two cases of the value $t(c)$, which are exactly the two cases of the position of $c$. If $c$ is rational then $t(c)>0$, and applying step 5.1 with the admissible choice $\epsilon :=t(c)$ gives ${\omega }_t(c)\le \max \{t(c),t(c)\}=t(c)$; with step 1.3 this gives ${\omega }_t(c)=t(c)$.

If $c$ is irrational then $t(c)=0$, and step 5.1 gives ${\omega }_t(c)\le \max \{0,\epsilon \}=\epsilon$ for every real $\epsilon >0$; since also ${\omega }_t(c)\ge 0$, an extended real that is $\le \epsilon$ for every positive real $\epsilon$ and $\ge 0$ must be $0$, so ${\omega }_t(c)=0=t(c)$.

Every real is rational or irrational and not both, so steps 6.1 and 6.2 establish claim 2 for every real $c$.

Claim 3 follows from claim 2: $t$ is continuous at $c$ exactly when ${\omega }_t(c)=0$, that is exactly when $t(c)=0$, that is exactly when $c$ is irrational. So the continuity set of $t$ is $\mathbb{R}\setminus \mathbb{Q}$ and its discontinuity set is $\mathbb{Q}$.