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[\cos\large\frac{\pi}{4}$$\sin x+\cos x\sin\large\frac{\pi}{4}]$
Step 2:
It is of the form $\sin(A+B)=\sin A\cos B+\cos A\sin B$
$\Rightarrow$ we get $\sqrt 2\sin(x+\large\frac{\pi}{4})$
At $x=a$ where $a\in R$
LHL=$\lim\limits_{\large x\to a^-}f(x)=\lim\limits_{\large a^-\to 0}\sqrt 2\sin(x+\large\frac{\pi}{4})$
$\qquad\qquad\qquad=\lim\limits_{\large h\to 0}\sqrt 2\sin(a-h+\large\frac{\pi}{4})$
$\qquad\qquad\qquad=\sqrt 2\lim\limits_{\large h\to 0}\sin(a+\large\frac{\pi}{4})$$\cos h-\cos(a+\large\frac{\pi}{4})$$\sin h]$
$\sin(A-B)=\sin A\cos B-\cos A\sin B$
$\qquad\qquad\qquad=\sqrt 2\sin(a+\large\frac{\pi}{4})$$\cos 0-\sqrt 2\cos(a+\large\frac{\pi}{4})$
$\qquad\qquad\qquad=\sqrt 2\sin[a+\large\frac{\pi}{4}]$
Step 3:
RHL=$\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large x\to a^+}\sqrt 2\sin(x+\large\frac{\pi}{4})$
$\qquad\qquad\qquad=\lim\limits_{\large h\to 0}\sqrt 2\sin(a+h+\large\frac{\pi}{4})$
$\qquad\qquad\qquad=\sqrt 2\lim\limits_{\large h\to 0}\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\qquad\qquad=\sqrt 2\sin(a+\large\frac{\pi}{4})$$\cos 0+\sin 0\cos(a+\large\frac{\pi}{4})$
Step 4:
$f(a)=\sqrt 2\sin(a+\large\frac{\pi}{4})$
LHL=RHL=f(a)
Hence $f(x)$ is continuous at all points.