Prove that if $\displaystyle f:R\rightarrow R$ is continuous and $\displaystyle f(x+y)=f(x)+f(y)$ for all $\displaystyle x,y\in R$, then there is a contstant $\displaystyle c\in R$, such that $\displaystyle f(x)=cx$, for all $\displaystyle x\in R$.

Here is my attempt:

Proof:

Let $\displaystyle f:R\rightarrow R$ and $\displaystyle f(x+y)=f(x)+f(y)$ for all $\displaystyle x,y\in R$, then there is a contstant $\displaystyle c\in R$, such that $\displaystyle f(x)=cx$, for all $\displaystyle x\in R$.

Uh?? Why do you write the conclusion you must prove??
Then, assume that f is NOT continuous.

Why would you do that? This is a given, so you cannot assume it isn't true! You could negate the conclusion and try to prove by contradiction,

but this ain't it.
Then, suppose some point $\displaystyle a$ is not a limit point for $\displaystyle f$. Then, since $\displaystyle f(a)=ac$

This is wrong: you can't construct ANY prove assuming that what you HAVE to prove is correct!!
, $\displaystyle f$ is defined at $\displaystyle a$. Now,let $\displaystyle a$ be a limit point for $\displaystyle f$. Then, $\displaystyle 0<|x-a|<\delta \Longrightarrow |f(x)-f(a)|<\epsilon$ for some$\displaystyle \epsilon>0$.

Then, $\displaystyle 0<|x-a|<\delta \Longrightarrow |f(x)-f(a)|<\epsilon$ is the same as $\displaystyle 0<|x-a|<\delta \Longrightarrow |cx-ca|<\epsilon$.

Now evaluate $\displaystyle |x-a|<\delta$. We have that $\displaystyle |x-a|<|c||x-a|=|cx-ca|<\epsilon$. Thus, $\displaystyle \delta =\epsilon$ works. However, we have shown that if $\displaystyle a$ is not a limit point at $\displaystyle f$ then $\displaystyle f$ is defined at $\displaystyle a$ and that if $\displaystyle a$ is a limit point of $\displaystyle f$ then $\displaystyle f(x)\longrightarrow f(a)$ as $\displaystyle x\longrightarrow a$. This is the definition of a continuous function so we have a contradiction. Thus, $\displaystyle f$ is continuous.

What a mess... ....this is NOT what you were asked to prove!!

You really must read much more carefully the problem and understand it, and then attempt to solve it.

Tonio
Any contructive critisism is welcome.