These germs make me sick! I need a "mini-crashcours" concerning the space of germs of continuous functions in order to solve an exercise which requires me to show that limits in this space aren't always unique.
We have introduced these ad-hoc to illustrate nonhausdorff spaces, so I dont have any intuition about germs and am rather frustrated because I don't have a clear picture about them in my head.
In our functional analysis course we have defined the set of germs  of continuous functions to be the $S/\sim$, where $$S:=\{ u:U\rightarrow \mathbb{R}:A\subseteq U\subseteq\mathbb{R}^n\ \text{open},u\big|_{U}\ \text{continuous}\}$$ and $A\subseteq\mathbb{R}^n$ is some fixed, closed set;  $\sim$ is the equivalence relation defined by $$u:U\rightarrow \mathbb{R} \sim v:V\rightarrow \mathbb{R} \Longleftrightarrow \exists W\subseteq\mathbb{R}^n:A\subseteq W, W \ \text{open}, \ W\subseteq(U\cap V),u\big|_W =v\big|_W.$$This set is endowed with the topology "generated by the final structure with respect to the collection of maps $(F_U)_U:C(U,\mathbb{R})\rightarrow S/\sim,\ f\mapsto [f]$."
Concerning the things in the "...":
 1) In which set do the $U$ from $(F_U)_U$ range ? The set of all open sets in $\mathbb{R}^n$ ?
2)With what topology is $C(U,\mathbb{R})$ endowed ? (For personal reasons asking the professor is unfortunately not an option) The topology of pointwise convergence, i.e. the product topology on $\prod_U \mathbb{R}$, restricted to the continuous functions ?

The Exercise I have to solve:
If $n=1$ and $A=\{0\}$ I have to show the limits aren't unique. 
Progress so far: Assuming that the answer to the questions above is "yes", the constant sequence of the equivalence classes of the constant $0$ function on some fixed open $U$, $[v]$, obviously converges in $S/\sim$ to $[v]$. But I have a hint that for continuous $u$ with $u(0)=0$ and $g_n(x):=u(x)h(nx)$ (where $h(x):=1$ for $|x|\geq2$ and $h(x)=0$ for $|x|\leq 1$ for $n\in \mathbb{N}$), we also have that the sequence $([g_n])_n$ converges to $[v]$.
But how do I prove the convergence of $([g_n])_n$?
 (Please note, that I have to prove this "elementary", since I don't have any theorems at my disposal that tell me what kind of convergence a final structure induces...)
 A: 
In which set do the $U$ from (F_U)_U range ? The set of all open sets in 
  $\mathbb{R}^n$ ?

Not quite. If $F_U(f) \in S/\sim$, then $f$ must be in $S$. That means
$U$ must be an open set containing $A$.

With what topology is $C(U, \mathbb{R})$ endowed ? The topology of pointwise  convergence, i.e. the product topology on $\prod_U \mathbb{R}$, restricted to the
   continuous functions ?

Generally, I would assume the compact-open topology. In this case it is enough
to assume that the topology is uniform convergence or any coarser topology.

But how do I prove the convergence of $([g_n])_n$?

The only thing you need to know about the final topology, is that it makes
every $F_U$ continuous. If you show that $g_n \to u$ in $C(U, \mathbb{R})$,
it follows immediately from that continuity that 
$[g_n] = F_U(g_n) \to F_U(u) = [u]$ in $S/\sim$.

Edit:
"Not Hausdorff" is a great understatement of how coarse the topology actually is.
The fact that a constant sequence can converge to a limit unequal to the terms
already proves that it is not T1, since $[u] \in \overline{\{[v]\}}$.
We can
go further than that by taking two arbitrary continuous functions $v$ and $u$
such that $v(0) = u(0)$ and defining $g_n(x) = v(x) + (u(x) - v(x))h_n(x)$.
By the same reasoning as before, we get $[u] \in \overline{\{[v]\}}$.
Since we made no assumptions on $u$ and $v$, other than that they agree at 0,
we can conclude that for any $c \in \mathbb{R}$ the subspace
$\{ [f] \in S/\sim \,\mid f(0) = c \}$ is indiscrete.
I have not checked this formally, but from there I would guess that $S/\sim$ is homeomorphic to the product of $\mathbb{R}$ and an indiscrete space.
That being said, I suspect that the objective of the exercise is 
precisely to demonstrate a way of proving things about a final
topology without having an explicit description in  terms of open sets,
as such a description  can be very complicated in more practical situations.
