$K>0$
$C_K$ : $z=(K-t)+it \,\,(0\leqq t \leqq K)$ (Straight Line on $\mathbb{C}$)
Prove that
(i) $|z|\geqq \dfrac{K}{\sqrt{2}}$ for $z\in C_K$
(ii) $\bigg|\dfrac{e^{iz}}{z}\bigg|\leqq \dfrac{\sqrt{2}}{K} e^{-t}$ for $z\in C_K$
(iii) $\lim_{K\to \infty} \displaystyle\int_{C_K} \dfrac{e^{iz}}{z} dz=0$
I could prove (i) and (ii), but I cannot prove (iii) .
\begin{align} \Bigg| \displaystyle\int_{C_K} \dfrac{e^{iz}}{z} dz \Bigg| &\leqq \sup_{z\in C_K} \Bigg| \dfrac{e^{iz}}{z} \Bigg| \cdot L(C_K) \\ &\leqq \dfrac{\sqrt{2}}{K}e^{-t}\cdot L(C_K) \\ &=\dfrac{\sqrt{2}}{K}e^{-t}\cdot \sqrt{K^2+K^2} \\ &=2e^{-t}. \end{align}
The term $K$ vanishes so it didn't work.
I would like you to give me some ideas.
Simply start to evaluate the integral, \begin{align} \left\lvert \int_{C_K} \frac{e^{iz}}{z} dz \right \rvert &=\left\lvert \int_0^K \frac{e^{(K-t)i-t}}{(K-t)+it} \cdot(-1+i) dt \right\rvert\\ &\leqslant\sqrt{2}\int_0^K \frac{e^{-t}}{\lvert (K-t) + it\rvert} dt \end{align} The denominator is bounded below by $\frac{1}{\sqrt 2} K$, so you obtain, \begin{align} \left\lvert \int_{C_K} \frac{e^{iz}}{z} dz \right \rvert &\leqslant\frac{2}{K} \int_0^K e^{-t} dt \\ &=\frac{2(1-e^{-K})}{K} \end{align} and this last term has limit zero as $K \to \infty$.
Comment on approach taken: the approximation using the length of $C_K$ and the maximum value of the integrand doesn't work because it cannot exploit the fact that $e^{iz}$ decreases rapidly along $C_K$. Moreover, in the answer you gave you still have a term $e^{-t}$, but the $t$ variable is only relevant inside the integral and therefore should never appear as part of the final estimate.