When I read the book "An introduction to Stochastic Differential Equations" by Evans, I am confused at Step 3 in the proof of existence Theorem of Ito diffusion.
Let $ \mathbf{X}$ be a ito diffusion defined as $$ \left\{\begin{aligned} d \mathbf{X} &=\mathbf{b}(\mathbf{X}, t) d t+\mathbf{B}(\mathbf{X}, t) d \mathbf{W} &(0 \leq t \leq T) \\ \mathbf{X}(0) &=\mathbf{X}_{0} \end{aligned}\right. . $$
To prove the existence of $ \mathbf{X}$, we use a sequence of $ \mathbf{X}^k$ which is defined as $$ \left\{\begin{aligned} \mathbf{X}^{0}(t) & :=\mathbf{X}_{0} \\ \mathbf{X}^{k+1}(t) & :=\mathbf{X}_{0}+\int_{0}^{t} \mathbf{b}\left(\mathbf{X}^{k}(s), s\right) d s+\int_{0}^{t} \mathbf{B}\left(\mathbf{X}^{k}(s), s\right) d \mathbf{W} \end{aligned}\right. . $$
In the proof, it make use of the fact that $$ \mathbf{X}^{k+1}(t)-\mathbf{X}^{k}(t) $$ is martingale. However, I can not see it directly. I know that $$ \int_{0}^{t} \mathbf{B}\left(\mathbf{X}^{k}(s), s\right) d \mathbf{W} $$ is a martingale. But I cann not see that the part $$ \int_{0}^{t} \mathbf{b}\left(\mathbf{X}^{k}(s), s\right)-\mathbf{b}\left(\mathbf{X}^{k-1}(s), s\right) d s $$ is also a martingale.
You have misunderstood. Evans is never claiming that $X^{k+1} - X^{k}$ is a martingale since he never applies the martingale inequality to this quantity. Rather, he splits $X^{k+1} - X^{k}$ into two terms; one of which is a martingale and the other of which can be bounded by the fact that $b$ is Lipschitz.
Writing out all the details very explicitly, we have $$X^{k+1} - X^{k} = \int_0^t b(X^{k}(s),s) - b(X^{k-1}(s),s) ds + \int_0^t B(X^{k}(s),s) - B(X^{k-1}(s),s) dW(s)$$ Hence, Evans is able to bound \begin{align} \max_{0 \leq t \leq T} |X^{k+1}(t) - X^{k}(t)|^2 \leq& \max_{0 \leq t \leq T} \bigg|2\bigg (\int_0^t b(X^{k}(s),s) - b(X^{k-1}(s),s) ds \bigg)^2 \\ & + 2\bigg( \int_0^t B(X^{k}(s),s) - B(X^{k-1}(s),s) dW(s) \bigg)^2 \bigg| \\ \leq & 2T \int_0^T |b(X^{k}(s),s) - b(X^{k-1}(s),s)|^2 ds \\ &+ 2 \max_{0\leq t \leq T} \bigg( \int_0^t B(X^{k}(s),s) - B(X^{k-1}(s),s) dW(s) \bigg)^2 \\ \leq& 2TL^2 \int_0^T |X^{k}(s) - X^{k-1}(s)|^2 ds \\ & + 2 \max_{0\leq t \leq T} \bigg( \int_0^t B(X^{k}(s),s) - B(X^{k-1}(s),s) dW(s) \bigg)^2 \end{align} where the last inequality uses the fact that $b$ is Lipschitz. Taking expectations yields \begin{align} \mathbb{E}[\max_{0 \leq t \leq T} |X^{k+1}(t) - X^{k}(t)|^2] \leq& 2TL^2 \mathbb{E}\bigg[\int_0^T |X^{k}(s) - X^{k-1}(s)|^2 ds \bigg] \\ & + 2\mathbb{E}\bigg[\max_{0\leq t \leq T} \bigg( \int_0^t B(X^{k}(s),s) - B(X^{k-1}(s),s) dW(s) \bigg)^2\bigg] \\ \leq& 2TL^2 \int_0^T \mathbb{E}[|X^{k}(s) - X^{k-1}(s)|^2] ds \\ & + 8 \int_0^T \mathbb{E}[|B(X^{k}(s),s) - B(X^{k-1}(s),s)|^2] ds \\ \leq& 2TL^2 \int_0^T \mathbb{E}[|X^{k}(s) - X^{k-1}(s)|^2] ds \\ &+ 8L^2 \int_0^T \mathbb{E}[|X^{k}(s) - X^{k-1}(s)|^2] ds \end{align} where the second inequality follows by applying Fubini's theorem to the first term and the martingale inequality only to the second term. The last line (which is the bound Evans needs) then follows since $B$ is Lipschitz also.