$f$ is continuous, $f(0)=1$, $f(m+n+1)=f(m)+f(n)$ for all real $m,n$. Show that $f(x)=1+x$ for all real $x$. In order to prove this, I had to back the proof up into a number of Lemmas. I am wondering if my method was reasonable, or if there was a more general or expedient path I might have taken? (I have left out all of the actual proofs with the exception of Lemma 6.)
Lemma 1:
For any $a\in\mathbb{R}$,   $$f(a)=1+a \implies f(-a)=1-a.$$
Lemma 2:
For any $n\in\mathbb{N}$, $\quad f(n)=1+n$.
Lemma 3:
For any $n\in\mathbb{N}$, $\quad f\left(\frac{1}{n}\right)=1+\frac{1}{n}$.
Lemma 4:
For any rational $\frac{m}{n}$, such that $0<m<n$, $\quad f\left(\frac{m}{n}\right)=1+\frac{m}{n}$.
Lemma 5:
For any $a\in\mathbb{Q}$, $\quad f(a)=1+a$.
Lemma 6:
For any irrational $x, \quad f(x)=1+x.$
Proof: Assume, for some irrational $x, \quad f(x)\neq 1+x$.
Say $f(x)>1+x$, then $$f(x)-(1+x)=t>0$$
and so, for any rational $a<x$ $$ f(x)-(1+a)>t.$$
However, by the definition of continuity, with $\epsilon=t$, for any rational satisfying $$x-\delta<a<x \quad\implies f(x)-(1+a)<t,$$
which is a contradiction. A similar argument shows that assuming $f(x)<1+x$ also leads to a contradiction, and it follows that $$f(x)=1+x.\quad \square$$
 A: If you make a translation $g(x) = f(x-1)$ the problem reduces to finding all continuous functions $g$, such that $g(1) = 1$, $g(x+y) = g(x) + g(y)$ for all real $x, y$. The second of these makes $g$ what is called an additive function; is a very standard functional equation which leads to $g(x)= g(1)x$ when any of the following additional assumptions is satisfied:


*

*$g$ is continuous at some point $x_0$ (you have that it is continuous everywhere, but it would have been enough to know it is continuous at any one arbitrary point)

*$g$ is bounded on some interval

*$g$ is monotonic on some interval
The proof, in general, proceeds exactly along the lines of those lemmas. But if you are allowed to use that a continuous additive function is always $g(x) = g(1)x$ as a theorem, you can "skip" all the lemmas.
A: By setting $m=0$, we see that $f(x+1)=f(x)+1$ for all real $x$. Since $f(0)=1$ we have $f(n)=n+1$ for all $n\in\Bbb Z$.
Lemma. $f(\frac n{2^m}) = 1+\frac n{2^m}$ for all $n,m\in\mathbb Z$.
Proof. If $m<0$ then the statement is simply a special case of $f(n)=n+1$. Suppose thus that $m\geq0$. I will use induction over $m$. Note that for $m=0$ the claim is still a special case of $f(n)=n+1$ and thus true.
Induction Step. Suppose that the Lemma is true for some $m\in\mathbb N\cup\{0\}$. Then for all $n\in\mathbb Z$,
\begin{split}
2\cdot f\left(\frac{n}{2^{m+1}}\right)&=f\left(\frac{n}{2^{m+1}}\right)+f\left(\frac{n}{2^{m+1}}\right)\\
&=f\left(\frac{n}{2^{m}}+1\right)\\
&=f\left(\frac{n}{2^{m}}\right)+1\\
&= \frac{n}{2^m}+2.
\end{split}
Between the first and second line I used the original assumption on $f$ and between the third and fourth line I used the inductive assumption.
It follows that $f\left(\frac{n}{2^{m+1}}\right)=\frac{n}{2^{m+1}}+1$. $\square$
Since $S:=\{\frac{n}{2^m}\mid n,m\in\mathbb Z\}$ is dense in $\Bbb R$, $f$ is continuous, and $f(x)=1+x$ for all $x\in S$, it follows that 
$$\bbox[5px,border:2px solid #C0A000]{f\equiv 1+x \text{ on } \mathbb R.}$$
