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Going off Fedor's answer, for $c$ at most $np+o(n)$, we may use the fact $P(X=c+1\mid p)=P(X=c\mid p)(1+o(1))$. It follows that for any $M>0$, $P(c\leq X\leq c+M-1\mid p)=(M+o(1))P(X=c\mid p)$. In other words, if we let $\varepsilon_1>0$, then there exists $N_{\varepsilon_1}>0$, such that when $n\geq N_{\varepsilon_1}$, $$ \left\vert\frac{P(c\leq X\leq c+M-1\mid p)}{P(X=c\mid p)}-M\right\vert <\varepsilon_1 $$ This implies $P(c\leq X\leq c+M-1\mid p)\geq P(X=c\mid p)(M-\varepsilon_1)$. Now let $\varepsilon_2>0$. Choosing $M>\varepsilon_1+\varepsilon_2^{-1}$, we have \begin{align*} \frac{P(X=c\mid p)}{P(X\geq\hat{c}^{(1)}\mid p)}&\leq \frac{P(X=c\mid p)}{P(c\leq X\leq c+M-1\mid p)}\\ &\leq \frac{P(X=c\mid p)}{P(X=c\mid p)(M-\varepsilon_1)}\leq \varepsilon_2 \end{align*}

Going off Fedor's answer, for $c$ at most $np+o(n)$, we may use the fact $P(X=c+1\mid p)=P(X=c\mid p)(1+o(1))$. It follows that for any $M>0$, $P(c\leq X\leq c+M-1\mid p)=(M+o(1))P(X=c\mid p)$. In other words, if we let $\varepsilon_1>0$, then there exists $N_{\varepsilon_1}>0$, such that when $n\geq N_{\varepsilon_1}$, $$ \left\vert\frac{P(c\leq X\leq c+M-1\mid p)}{P(X=c\mid p)}-M\right\vert <\varepsilon_1 $$ This implies $P(c\leq X\leq c+M-1\mid p)\geq P(X=c\mid p)(M-\varepsilon_1)$. Now let $\varepsilon_2>0$. Choosing $M>\varepsilon_1+\varepsilon_2^{-1}$, we have \begin{align*} \frac{P(X=c\mid p)}{P(X\geq c\mid p)}&\leq \frac{P(X=c\mid p)}{P(c\leq X\leq c+M-1\mid p)}\\ &\leq \frac{P(X=c\mid p)}{P(X=c\mid p)(M-\varepsilon_1)}\leq \varepsilon_2 \end{align*}

Going off Fedor's answer, for $c$ at most $np+o(n)$, we may use the fact $P(X=c+1\mid p)=P(X=c\mid p)(1+o(1))$. It follows that for any $M>0$, $P(c\leq X\leq c+M-1\mid p)=(M+o(1))P(X=c\mid p)$. In other words, if we let $\varepsilon_1>0$, then there exists $N_{\varepsilon_1}>0$, such that when $n\geq N_{\varepsilon_1}$, $$ \left\vert\frac{P(c\leq X\leq c+M-1\mid p)}{P(X=c\mid p)}-M\right\vert <\varepsilon_1 $$ This implies $P(c\leq X\leq c+M-1\mid p)\geq P(X=c\mid p)(M-\varepsilon_1)$. Now let $\varepsilon_2>0$. Choosing $M>\varepsilon_1+\varepsilon_2^{-1}$, we have \begin{align*} \frac{P(X=c\mid p)}{P(X\geq c\mid p)}&\leq \frac{P(X=c\mid p)}{P(c\leq X\leq c+M-1\mid p)}\\ &\leq \frac{P(X=c\mid p)}{P(X=c\mid p)(M-\varepsilon_1)}\leq \varepsilon_2 \end{align*}

Going off Fedor's answer, for $c$ at most $np+o(n)$, we may use the fact $P(X=c+1\mid p)=P(X=c\mid p)(1+o(1))$. It follows that for any $M>0$, $P(c\leq X\leq c+M-1\mid p)=(M+o(1))P(X=c\mid p)$. In other words, if we let $\varepsilon_1>0$, then there exists $N_{\varepsilon_1}>0$, such that when $n\geq N_{\varepsilon_1}$, $$ \left\vert\frac{P(c\leq X\leq c+M-1\mid p)}{P(X=c\mid p)}-M\right\vert <\varepsilon_1 $$ This implies $P(c\leq X\leq c+M-1\mid p)\geq P(X=c\mid p)(M-\varepsilon_1)$. Now let $\varepsilon_2>0$. Choosing $M>\varepsilon_1+\varepsilon_2^{-1}$, we have \begin{align*} \frac{P(X=c\mid p)}{P(X\geq\hat{c}^{(1)}\mid p)}&\leq \frac{P(X=c\mid p)}{P(c\leq X\leq c+M-1\mid p)}\\ &\leq \frac{P(X=c\mid p)}{P(X=c\mid p)(M-\varepsilon_1)}\leq \varepsilon_2 \end{align*}

Going off Fedor's answer, for $c$ at most $np+o(n)$, we may use the fact $P(X=c+1\mid p)=P(X=c\mid p)(1+o(1))$. It follows that for any $M>0$, $P(c\leq X\leq c+M-1\mid p)=(M+o(1))P(X=c\mid p)$. In other words, if we let $\varepsilon_1>0$, then there exists $N_{\varepsilon_1}>0$, such that when $n\geq N_{\varepsilon_1}$, $$ \left\vert\frac{P(c\leq X\leq c+M-1\mid p)}{P(X=c\mid p)}-M\right\vert <\varepsilon_1 $$ This implies $P(c\leq X\leq c+M-1\mid p)\geq P(X=c\mid p)(M-\varepsilon_1)$. Now let $\varepsilon_2>0$. Choosing $M>\varepsilon_1+\varepsilon_2^{-1}$, we have \begin{align*} \frac{P(X=c\mid p)}{P(X\geq\hat{c}^{(1)}\mid p)}&\leq \frac{P(X=c\mid p)}{P(c\leq X\leq c+M-1\mid p)}\\ &\leq \frac{P(X=c\mid p)}{P(X=c\mid p)(M-\varepsilon_1)}\leq \varepsilon_2 \end{align*}

Going off Fedor's answer, for $c$ at most $np+o(n)$, we may use the fact $P(X=c+1\mid p)=P(X=c\mid p)(1+o(1))$. It follows that for any $M>0$, $P(c\leq X\leq c+M-1\mid p)=(M+o(1))P(X=c\mid p)$. In other words, if we let $\varepsilon_1>0$, then there exists $N_{\varepsilon_1}>0$, such that when $n\geq N_{\varepsilon_1}$, $$ \left\vert\frac{P(c\leq X\leq c+M-1\mid p)}{P(X=c\mid p)}-M\right\vert <\varepsilon_1 $$ This implies $P(c\leq X\leq c+M-1\mid p)\geq P(X=c\mid p)(M-\varepsilon_1)$. Now let $\varepsilon_2>0$. Choosing $M>\varepsilon_1+\varepsilon_2^{-1}$, we have \begin{align*} \frac{P(X=c\mid p)}{P(X\geq c\mid p)}&\leq \frac{P(X=c\mid p)}{P(c\leq X\leq c+M-1\mid p)}\\ &\leq \frac{P(X=c\mid p)}{P(X=c\mid p)(M-\varepsilon_1)}\leq \varepsilon_2 \end{align*}