Let $(\Bbb R, \mathcal A_{\Bbb R}^*,\overline{\lambda})$ be the complete lebesgue-measure space. Let $f:\Bbb R\to \Bbb R$ be an additive function and also Lebesgue-measurable:
$$\text{I want to prove that:}\;\;f(x)=f(1)\ x\;\;\forall x\in\Bbb R$$
Proof:
So first I prove some clearly results.
Lemma 1: $\;f(x-y)=f(x)-f(y)\;\forall x,y\in\Bbb R$
Let $x,y\in\Bbb R$ so, since $f$ is additive, we get that:
$$f(0)=f(0+0)=f(0)+f(0)\Rightarrow\ f(0)=0\\
\Rightarrow\ 0=f(0)=f(x+(-x))=f(x)+f(-x)\Leftrightarrow\ -f(x)=f(-x)$$
thereby $\;f(x-y)=f(x)+f(-y)=f(x)-f(y).$
Lemma 2: $\;r\ f(x)=f(rx)\;\;\forall r\in\Bbb Q\;\forall x\in\Bbb R$
Let $x\in\Bbb R$ we state that $\;n\ f(x)=f(nx)\;\forall n\in\Bbb N$. Because (by induction over n) for $n=2$ is clear that $f(2x)=f(x+x)=f(x)+f(x)=2f(x)$, if we suppose that $f(nx)=nf(x)$ follows for $n$ we get that $f((n+1)x)=f(nx+x)=f(nx)+f(x)=nf(x)+f(x)=(n+1)f(x)$.
Then let $\;m=-n\;\;\forall n\in\Bbb N$ thus $\;f(mx)=f(-nx)=-nf(x)=mf(x)$ and hence $f(zn)=zf(x)\;\;\forall z\in\Bbb Z\Rightarrow\ r=\frac{n}{m}$ with $m,n\in\Bbb Z$, $(n,m)=1$ and $m\ne 0$
By last, let $r\in\Bbb Q\setminus\Bbb Z$. So, since $f(mx)=mf(x),$ by taking $x=\frac{y}{m}$ and $q=\frac{1}{m}$ we get that $f(y)=\frac{1}{q}f\big(\frac{y}{m}\big)\Leftrightarrow\ qf(y)=f\big(\frac{y}{m}\big)\Leftrightarrow\ \frac{1}{m}f(y)=f\big(\frac{y}{m}\big)$ and thus $f(rx)=f\big(\frac{n}{m} x\big)=nf\big(\frac{1}{m}x\big)=\frac{n}{m}f(x)=rf(x)$.
Now I want to prove that $f$ is continuous:
So, it's clear that:
$$\bigcup_{m>0}\{x\in\Bbb R:|f(x)|\le m\}=\Bbb R\ne\emptyset$$
Let $A_m=\{x\in\Bbb R:|f(x)|\le m\}$, then for each fixed $t\in\Bbb R\;\exists\ m_t>0\;$ s.t. $\;t\in A_{m_t}\;$ so $\;\overline{\lambda}(A_{m_t})>0$.(where $A_m\in\mathcal A_{\Bbb R}^*$ follows since $f^{-1}([-m,m])=A_m$ and $f$ is measurable)
So, by Steinhaus Theorem, there $\exists\ \delta>0\;$ s.t. if $\;|x|<\delta$ we get that:
$$\overline{\lambda}\big(A_{m_t}\cap (A_{m_t}+x)\big)>0\\ \text{and that}\;\; (-\delta,\delta)\subset \{a-a':a,a'\in A_{m_t}\}=:\ A_{m_t}-A_{m_t}$$
Thereby $\forall x\in (-\delta,\delta)\;\;\exists\ a,a'\in A_{m_t}\;$ s.t. $\; x=a-a'\Rightarrow\ f(x)=f(a-a')=f(a)-f(a')\Rightarrow\ |f(x)|=|f(a)-f(a')|\le m_t+m_t=2m_t$.
Namely $\forall x\in (-\delta,\delta)\Rightarrow\ |f(x)|\le 2m_t$ so taking $\delta^*=\frac{\delta}{2^{n+1}}$ for $\; n\in\Bbb N\;\;$ we get that:
$$\text{if}\;\; |x-y|<\delta^*\Rightarrow\ 2^{n+1}|x-y|<\delta\Rightarrow\ |2^{n+1}(x-y)|<\delta\\ \Rightarrow\Big|f\big(2^{n+1}(x-y)\big)\Big|\le 2m_t$$
thus (by lemma 1 and 2)
$$|f(x)-f(y)|=|f(x-y)|=\frac{1}{2^{n+1}}\Big|f\big(2^{n+1}(x-y)\big)\Big|\le \frac{2m_t}{2^{n+1}}=\frac{m_t}{2^n}\\ \Rightarrow\ |f(x)-f(y)|\le \frac{m_t}{2^n}\to 0,\;\;\ n\to\infty $$
therefore $f$ is continuous.
So at last, let $\epsilon>0$:
If $f(1)=0$, $f(x)=xf(1)$ follows straightforward.
If $f(1)\ne 0$, we take $\delta_{\epsilon}=\frac{\epsilon}{2|f(1)|}$ so if $|x|<\delta_{\epsilon}$ then
$$|f(x)-xf(1)|\le |f(x)|+|x||f(1)|$$
where $|f(x)|<\frac{\epsilon}{2}$ by the particular continuity of $f$ at $0$, so
$$|f(x)-xf(1)|\le |f(x)|+|x||f(1)|<\frac{\epsilon}{2}+\delta_{\epsilon}|f(1)|=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
hence $f(x)=xf(1)\;\;\forall x\in\Bbb R.$
Hint: I cannot follow easily your proof. Now, I remember another proof that used a certain trick. One should consider the function $\chi(x) = e^{i f(x)}$ whose values are complex of absolute value $1$. Say we want to show that $\chi$ is continuous at $0$ ( from this the continuity of $f$ follows easily). Consider a finite interval $[a,b]$. We have \begin{eqnarray} \int_a^b \chi(x+y) d x = \chi(y) \int_a ^b \chi(x) d x \end{eqnarray} Now for locally $L^1$ functions like $\chi$ we have $$\lim_{y\to 0} \int_a^b \chi(x+y) d x = \int_a^b \chi(x) d x $$ Now choose a convenient $[a,b]$ so that the integral $\int_a^b \chi(x) d x$ is not zero.