real analysis - If $f(x+y)=f(x)+f(y) ,forall;x,yinBbb{R}$, then if $f$ is continuous at $0$, then it is continuous on $Bbb{R}.$

I know that this question has been asked here before but I want to use a different approach. Here is the question.

A function $f:\Bbb{R}\to\Bbb{R}$ is such that
\begin{align} f(x+y)=f(x)+f(y) ,\;\;\forall\;x,y\in\Bbb{R}\qquad\qquad\qquad(1)\end{align}
I want to show that if $f$ is continuous at $0$, it is continuous on $\Bbb{R}.$

MY WORK

Since $(1)$ holds for all $x\in \Bbb{R},$ we let \begin{align} x=x-y+y\end{align}
Then,

\begin{align} f(x-y+y)=f(x-y)+f(y)\end{align}
\begin{align} f(x-y)=f(x)-f(y)\end{align}
Let $x_0\in \Bbb{R}, \;\epsilon>$ and $y=x-x_0,\;\;\forall\,x\in\Bbb{R}.$ Then,
\begin{align} f(x-(x-x_0))=f(x)-f(x-x_0)\end{align}
\begin{align} f(x_0)=f(x)-f(x-x_0)\end{align}
\begin{align} f(y)=f(x_0)-f(x)\end{align}

HINTS BY MY PDF:

Let $x_0\in \Bbb{R}, \;\epsilon>$ and $y=x-x_0,\;\;\forall\,x\in\Bbb{R}.$ Then, show that \begin{align} \left|f(x_0)-f(x)\right|=\left|f(y)-f(0)\right|\end{align}

Using this equation and the continuity of $f$ at $0$, establish properly that
\begin{align}\left|f(y)-f(0)\right|<\epsilon,\end{align}
in some neighbourhood of $0$.

My problem is how to put this hint together to complete the proof. Please, I need assistance, thanks!

Answer

We want to show that

$$\forall \epsilon>0, \exists r>0:|x-y|

But $f(x)-f(y)=f(x-y)$ because $f(y)+f(x-y)=f(y+(x-y))=f(x)$ as you have noticed.

Now, take $u=x-y$. By continuity at $0$, we can write:

$$\forall \epsilon>0, \exists r>0:|u-0|

It's easy to see that $f(0)=0$, because $f(0)=f(0+0)=f(0)+f(0)$. Hence

$$\forall \epsilon>0, \exists r>0:|(x-y)-0| $$\forall \epsilon>0, \exists r>0:|x-y|
Hence, $f$ is continuous at any $y \in \mathbb{R}$.