Reason for the integer case and the rational case to be solved differently 
Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
  for all real $m, n$. 
  Show that $f(x) = 1 + x$ for all real numbers $x$.

This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;

first prove this for integer $x$, then for rational $x$, then finally for real $x.$

The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?
Any help would be much appreciated. 

Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.
Continous Additive functions are linear
Coming back to your Orignal Problem
Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
Keep it aside for a while.
By the functional equation  given $f(y)=f({\color{red}{0}}+{\color{blue}{y-1}}+1)=f({\color{red}{0}})+f({\color{blue}{y-1}})$
Hence $f(y)=1+f(y-1)$
Further consider $f(x+y)$
$$f({\color{red}{x}}+{\color{blue}{y}})=f({\color{red}{x}}+{\color{blue}{y-1}}+1)$$
$$f({\color{red}{x}}+{\color{blue}{y}})=f({\color{red}{x}})+f({\color{blue}{y-1}})$$
Hence we have $f(x+y)=f(x)+f(y)-1$
This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$ 
Hence $g(x+y)=g(x)+g(y)$
Hence $g(x)=g(1)x$
Which implies $f(x)-1=g(1)x$
Hence $f(x)=1+(f(x)-1)x$
Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 \times 1=2$
We conclude $f(x)=1+x$
You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above
