If $f$ is continuous on $\mathbb R$ then $\exists c\in\mathbb R: f(x)=c$ has only one solution I have to prove that there is no continuous function $f: \mathbb{R} \to \mathbb{R}$ such that, for each $c \in \mathbb{R}$ the equation $f(x)=c$ has exactly two solutions.
My attempt:
We have that since $f$ is continuous we can define a sequence $\{x_n\}$ in the reals such that $\{x_n\} \to x$ then $\{f(x_n)\} \to f(x)=c$.
Now consider a sequence $\{y_n\}$ in the reals too, such that $\{y_n\} \to x$ then $\{f(y_n)\} \to f(x)$ but because of the uniqness of the limit the result follows.
Am I right? thank you    
I think for the second question What happens then between u and v ? the function is constant isn't otherwise we repeat the value of f(α)=f(β)=e three times and we are not allowed to do this, am I right @Herbert Quain 
 A: Okay, I wrote the solution but I omitted some key points : try to find them by yourself tomorrow. If you understand the preceding arguments you should not be blocked, otherwise ask for help. 
When you have a problem about continuous functions and solutions of equations $f(t) = c$, you should immediately think about the intermediate value theorem : the proof to come only consists in working with this theorem.
Let's suppose that $f$ is continuous and such that $f(x) = c$ has exactly two solutions. Choose $c$ and suppose $f(x) = f(y)=c$, with $x < y$. 
$f$ being continuous between $x$ and $y$, she attains her maximum on this interval at a point $v$. 
1) Suppose $x<v<y$ and write $f(v) = M$.Then, $M>c$ (why ?) and there is another point $v$ such that $f(u) = M$. Suppose $u$ is not in $]x,y[$, say $u<x<v<y$.
Consider a real $e$ such that $c <e <M$.Then, if $f$ is continuous, by the intermediate value theorem, there is at least one solution of $f(t)=e$ in the following intervals :  $]u,x[$,$]x,v[$,$]v,x[$. This is wrong by hypothesis. 
So we must have  $x<u<v<y$. Let's sum up the thing using once again the intermediate value theorem : there is one point $\alpha$ in $]x,u[$ such that $f(\alpha) = e$. There is also a point $\beta$ in $]v,y[$ such that $f(\beta) = e$. What happens then between $u$ and $v$ ? 
Conclusion : case 1) is impossible.
2) Therefore we must have $x=v$ or $y=v$, say $x=v$. Then, for every $z \in [x,y]$, $f(z) \leq c$ (why ?). But $f$ being continuous on the interval $[x,y]$, she also attains her minimum at a point $v'$ with $f(v')=m$. We can't have $m=c$ (why ?), so we must have $c>m$. Now apply 1) to the function $-f$ (why is it possible ?). 
Both cases are impossible : the situation is absurd.
A: We will aim to contradiction. Let $c_0$ be the value of $f(0)$. Then there is another value $x_0$ for which $f(x_0)=c_0$. If $c_0$ is taken exactly two times, then on the interval $(0,x_0)$ value $c_0$ is never taken, so by intermediate value theorem we have that on this interval the values of the function are either always less than $c_0$ or greater than it. Assume the former. Let $c_{min}$ be local minimum of $f$ on $[0,x_0]$, and let $x_{min}$ attan this value. From continuity we can deduce that on $(0,x_{min})$ the function is strictly decreasing, and on $(x_{min},x_0)$ it's increasing, because otherwise some value would be taken more than twice. But from assumption there is another value of $x$ for which $f(x)=x_{min}$, necessarilly outside the interval. Suppose that this $x$ is greater than $x_0$. Then on interval $(x_0,x)$ function takes all values between $c_0$ and $c_min$, which are also taken on intervals $(0,x_{min})$ and $(x_{min},x_0)$, so these values are taken thrice, contradiction.
A: (1) if $f$ has a local maximum (or minimum) the 2-point fiber condition implies this must be a global maximum (minimum). but then $f$ is not surjective as required.
(2) otherwise, since $f$ cannot be constant on an interval, it must be strictly monotonic, hence a bijection. in contradiction to the two-point fiber condition.
i hope this encapsulates our intuition in a straightforward way. the challenge, for this approach, is to find the most elegant and lucid way of applying the intermediate-value theorem to establish (1)
