Step 1:
$f(x)=\sin x-\cos x$
Multiply and divide by $\sqrt 2$
$\qquad=\sqrt 2[\large\frac{1}{\sqrt 2}$$\sin x-\large\frac{1}{\sqrt 2}$$\cos x]$
$\qquad=\sqrt 2[\sin x\cos\large\frac{\pi}{4}$$-\cos x\sin\large\frac{\pi}{4}]$
The above equation is of the form $\sin(a-b)=\sin a\cos b-\cos a\sin b$
$\Rightarrow \sqrt 2\sin(x-\large\frac{\pi}{4})$
Step 2:
At $x=a$ where $a\in R$
LHL=$\lim\limits_{\large x\to a^-}\sqrt 2\sin(x-\large\frac{\pi}{4})$
$\qquad=\lim\limits_{h\to 0}\sqrt 2\sin(a-h-\large\frac{\pi}{4})$
$\qquad=\lim\limits_{h\to 0}\sqrt 2[\sin(a-\large\frac{\pi}{4})$$\cos h-\cos(a-\large\frac{\pi}{4})$$\sin h$
$\sin(A-B)=\sin A\cos B-\cos A\sin B$
$\qquad=\sqrt 2[\sin(a-\large\frac{\pi}{4})$$\cos 0-\sqrt 2\cos(a-\large\frac{\pi}{4})$$\sin 0$
$\qquad=\sqrt 2\sin(a-\large\frac{\pi}{4})$
Step 3:
RHL=$\lim\limits_{\large x\to a^+}\sqrt 2\sin(x-\large\frac{\pi}{4})$
$\qquad=\lim\limits_{h\to 0}\sqrt 2\sin(a+h-\large\frac{\pi}{4})$
$\qquad=\lim\limits_{h\to 0}\sqrt 2[\sin(a-\large\frac{\pi}{4})$$\cos h+\cos(a-\large\frac{\pi}{4})$$\sin h$
$\qquad=\sqrt 2[\sin(a-\large\frac{\pi}{4})$$\cos 0-\cos(a-\large\frac{\pi}{4})$$\sin 0]$
$\qquad=\sqrt 2\sin(a-\large\frac{\pi}{4})$
Step 4:
Also $f(a)=\sqrt 2\sin(a-\large\frac{\pi}{4})$
$\therefore$ LHL =RHL=f(a).
Hence $f(x)$ is continuous at all points.