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First here, there is a typo in the Dvoretzky paper: there must be $-1+\frac{1}{2}t^2E[X_{n,k}^2|F_{n,k-1}]$ instead of $-1-\frac{1}{2}t^2E[X_{n,k}^2|F_{n,k-1}]$ there. Otherwise, the inequality will not be true in general. Indeed, let, for brevity, $X:=X_{n,k}$, $F:=F_{n,k-1}$, $E_F Z:=E(Z|F)$, and $c:=\epsilon$. Suppose, e.g., that $X$ is independent of $F$, $P(X=1)=P(X=-1)=1/2$, and $c=1$. Then the erroneous inequality becomes \begin{equation} |\cos t-1-t^2/2|\le\tfrac16\,|t|^3, \end{equation} which is false for small enough $|t|$, since $|\cos t-1-t^2/2|\sim t^2$ as $t\to0$.

So, we actually need to show that $$|E_F\exp(itX)-1+\tfrac12\,t^2E_F X^2|\le \tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c} $$ given that $E_F X=0$.

By Taylor's theorem with the integral form of the remainder, \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|x|^3/6 \tag{1} \end{equation*} and \begin{equation*} |e^{ix}-1-ix|\le x^2/2 \end{equation*} for real $x$. The latter inequality also implies \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|e^{ix}-1-ix|+|(ix)^2/2|\le x^2/2+x^2/2=x^2. \tag{2} \end{equation*}

Let $r(x):=e^{ix}-1-ix-(ix)^2/2$ and write, in view of (1) and (2): \begin{multline*} |E_F\exp(itX)-1+\tfrac12\,t^2E_F X^2| =|E_F r(tX)|\le|E_F r(tX)\,1_{|X|\le c}|+|E_F r(tX)\,1_{|X|<c}| \\ \le\tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c}, \end{multline*} as desired.

First here, there is a typo in the Dvoretzky paper: there must be $-1+\frac{1}{2}t^2E[X_{n,k}^2|F_{n,k-1}]$ instead of $-1-\frac{1}{2}t^2E[X_{n,k}^2|F_{n,k-1}]$ there. Otherwise, the inequality will not be true in general. Indeed, let, for brevity, $X:=X_{n,k}$, $F:=F_{n,k-1}$, $E_F Z:=E(Z|F)$, and $c:=\epsilon$. Suppose, e.g., that $X$ is independent of $F$, $P(X=1)=P(X=-1)=1/2$, and $c=1$. Then the erroneous inequality becomes \begin{equation} |\cos t-1-t^2/2|\le\tfrac16\,|t|^3, \end{equation} which is false for small enough $|t|$, since $|\cos t-1-t^2/2|\sim t^2$ as $t\to0$.

So, we actually need to show that $$|E_F\exp(itX)-1+\tfrac12\,t^2E_F X^2|\le \tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c} $$ given that $E_F X=0$.

By Taylor's theorem with the integral form of the remainder, \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|x|^3/6 \tag{1} \end{equation*} and \begin{equation*} |e^{ix}-1-ix|\le x^2/2 \end{equation*} for real $x$. The latter inequality also implies \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|e^{ix}-1-ix|+|(ix)^2/2|\le x^2/2+x^2/2=x^2. \tag{2} \end{equation*}

Let $r(x):=e^{ix}-1-ix-(ix)^2/2$ and write, in view of (1) and (2): \begin{multline*} |E_F\exp(itX)-1+\tfrac12\,t^2E_F X^2| =|E_F r(tX)|\le|E_F r(tX)\,1_{|X|\le c}|+|E_F r(tX)\,1_{|X|>c}| \\ \le\tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c}, \end{multline*} as desired.

By Taylor's theorem with the integral form of the remainder, \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|x|^3/6 \tag{1} \end{equation*} and \begin{equation*} |e^{ix}-1-ix|\le x^2/2 \end{equation*} for real $x$. The latter inequality also implies \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|e^{ix}-1-ix|+|(ix)^2/2|\le x^2/2+x^2/2=x^2. \tag{2} \end{equation*}

Let now $X:=X_{n,k}$, $F:=F_{n,k-1}$, $E_F Z:=E(Z|F)$, and $c:=\epsilon$. We need to show that $$|E_F\exp(itX)-1-\tfrac12\,t^2E_F X^2|\le \tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c} $$ given that $E_F X=0$.

Let $r(x):=e^{ix}-1-ix-(ix)^2/2$ and write, in view of (1) and (2): \begin{multline*} |E_F\exp(itX)-1-\tfrac12\,t^2E_F X^2| =|E_F r(itX)|\le|E_F r(itX)\,1_{|X|\le c}|+|E_F r(itX)\,1_{|X|<c}| \\ \le\tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c}, \end{multline*} as desired.

First here, there is a typo in the Dvoretzky paper: there must be $-1+\frac{1}{2}t^2E[X_{n,k}^2|F_{n,k-1}]$ instead of $-1-\frac{1}{2}t^2E[X_{n,k}^2|F_{n,k-1}]$ there. Otherwise, the inequality will not be true in general. Indeed, let, for brevity, $X:=X_{n,k}$, $F:=F_{n,k-1}$, $E_F Z:=E(Z|F)$, and $c:=\epsilon$. Suppose, e.g., that $X$ is independent of $F$, $P(X=1)=P(X=-1)=1/2$, and $c=1$. Then the erroneous inequality becomes \begin{equation} |\cos t-1-t^2/2|\le\tfrac16\,|t|^3, \end{equation} which is false for small enough $|t|$, since $|\cos t-1-t^2/2|\sim t^2$ as $t\to0$.

So, we actually need to show that $$|E_F\exp(itX)-1+\tfrac12\,t^2E_F X^2|\le \tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c} $$ given that $E_F X=0$.

By Taylor's theorem with the integral form of the remainder, \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|x|^3/6 \tag{1} \end{equation*} and \begin{equation*} |e^{ix}-1-ix|\le x^2/2 \end{equation*} for real $x$. The latter inequality also implies \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|e^{ix}-1-ix|+|(ix)^2/2|\le x^2/2+x^2/2=x^2. \tag{2} \end{equation*}

Let $r(x):=e^{ix}-1-ix-(ix)^2/2$ and write, in view of (1) and (2): \begin{multline*} |E_F\exp(itX)-1+\tfrac12\,t^2E_F X^2| =|E_F r(tX)|\le|E_F r(tX)\,1_{|X|\le c}|+|E_F r(tX)\,1_{|X|<c}| \\ \le\tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c}, \end{multline*} as desired.

This inequality follows immediately from the inequalities $$|e^{ix}-1-ix-(ix)^2/2|\le|x|^3/6 \tag{1}$$ and $$|e^{ix}-1-ix|\le x^2/2 \tag{2}$$ for real $x$, as the latter inequality also implies $$|e^{ix}-1-ix-(ix)^2/2|\le|e^{ix}-1-ix|+|(ix)^2/2|\le x^2/2+x^2/2=x^2.$$

In turn, inequalities (1) and (2) follow immediately from Taylor's theorem with the integral form of the remainder.

By Taylor's theorem with the integral form of the remainder, \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|x|^3/6 \tag{1} \end{equation*} and \begin{equation*} |e^{ix}-1-ix|\le x^2/2 \end{equation*} for real $x$. The latter inequality also implies \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|e^{ix}-1-ix|+|(ix)^2/2|\le x^2/2+x^2/2=x^2. \tag{2} \end{equation*}

Let now $X:=X_{n,k}$, $F:=F_{n,k-1}$, $E_F Z:=E(Z|F)$, and $c:=\epsilon$. We need to show that $$|E_F\exp(itX)-1-\tfrac12\,t^2E_F X^2|\le \tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c} $$ given that $E_F X=0$.

Let $r(x):=e^{ix}-1-ix-(ix)^2/2$ and write, in view of (1) and (2): \begin{multline*} |E_F\exp(itX)-1-\tfrac12\,t^2E_F X^2| =|E_F r(itX)|\le|E_F r(itX)\,1_{|X|\le c}|+|E_F r(itX)\,1_{|X|<c}| \\ \le\tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c}, \end{multline*} as desired.