Let $$f_{n}(x)=\frac{1+x}{1+\exp(nx)},\qquad n\in \mathbb{N}$$ be defined on $\mathbb{R}$. Prove that this is not uniformly convergent on the interval, that includes zero.
Without proving it, first we see that $$ f(x)=\lim_{n\to\infty}\frac{1+x}{1+\exp(nx)}=\begin{cases} 0 & \text{ if } x>0 \\ \frac{1}{2} & \text{ if } x=0 \\ 1+x & \text{ if } x<0 \end{cases} $$ so the pointwise limit $f(x)$ is continous for all $x\in \mathbb{R}\setminus \lbrace 0\rbrace$. That is, it is not continous on $0$. Thus $\lbrace f_{n}\rbrace$ is not uniformly convergent on $\mathbb{R}$. This is what I have read about the uniformly convergene of the sequence of function in a Theorem as shown below (translated from Norwegian)
Let $f$ and $f_{1},f_{2},f_{3},\dots$ be function defined on $A$. Suppose that $f_{1},f_{2},f_{3},\dots$ are continous and that $\lbrace f_{n}\rbrace$ uniformly converges to $f$ on $A$. Then $f$ is continous on $A$.
Now I have to prove it (look at the bold text) - which is my problem I don't know how and where to start. I think I would prove it by a contradiction. Can you help me?
Note that I don't have to prove the theorem. I only need to prove that this sequence of function isn't uniformly convergent.
The theorem you mention in your question is classical result, and the proof is actually quite simple (see the uniform limit theorem). Even if you don't end up using it, I really recommend reading it, as it makes use of a trick you can use in many proofs regarding convergence of functions.
Now, using this theorem to prove what you want is easy: As you mention in one of your comments you can use the contrapositive of the result:
Since $f_n$ is continuous for all $n$, the result immediately follows.
If you don't want to use the theorem, I would proceed along those lines: Recall that $g_n$ converges uniformly to $g$ is and only if $\|g_n-g\|_{\infty}\to0$, where $\|h\|_{\infty}=\sup\{h(x):x\in I\}$ for every function $h:I\subset \mathbb R\to\mathbb R$.
If you've studied uniform convergence, you've probably seen that if $f_n$ converges uniformly to some function $f$, then $f_n$ converges pointwise to the same function $f$. Therefore, since pointwise limits are unique, if $f_n$ converges uniformly, it must do so towards the piecewise function $f$ you've introduced in your question. Thus, to prove that $f_n$ doesn't converge uniformly at all, all we need to do is prove that it doesn't converge uniformly to $f$.
So let $(a,b)\subset\mathbb R$ be an interval that includes zero, and let $\epsilon>0$ be arbitrary. Fix $n\in\mathbb N$. Since $f_n$ is continuous and $f_n(0)=1/2$, then for every $\epsilon>0$, there exists $\delta_n>0$ such that if $|x-0|<\delta_n$, then $|f_n(x)-f_n(0)|=|f_n(x)-1/2|<\epsilon$, or, in other words, $f_n(x)\in(1/2-\epsilon,1/2+\epsilon)$, which implies that $f_n(x)>1/2-\epsilon$.
Let $c_n=\min\{\delta_n,b\}$. Then, we have that $c_n>0$ (since $\delta_n,b>0$), and that for any $x\in(0,c_n)$, \begin{align*} |f_n(x)-f(x)|&=|f_n(x)-0|&\text{(since $x>0$)}\\ &=f_n(x)&\text{(since $f_n(x)>0$ if $x\geq0$)}\\ &>\epsilon-1/2. \end{align*} Therefore, since $n\in\mathbb N$ was arbitrary, we conclude that $$\|f_n-f\|_{\infty}\geq\epsilon-1/2\text{ for every }n\in\mathbb N.$$ If you pick $\epsilon<1/2$, one can easily see that this implies the desired result.