If $\lim \limits_{x \to a} f(x)$ exists and $\lim \limits_{x \to a} g(x)$ does not exist, then $\lim \limits_{x \to a} (f+g)(x)$ does not exist. Could somone verify this proof? 

If $\lim \limits_{x \to a} f(x)$ exists and $\lim \limits_{x \to a}
g(x)$ does not exist, then $\lim \limits_{x \to a} (f+g)(x)$ does not
  exist. 
$$\forall m \ \exists \epsilon \ \forall \delta>0 \  \exists x : |x - a| \implies |g(x) - m| > \epsilon$$
Let $\overline{x}$ be the minimum of such $x$. Then if $m = m' -f(\overline{x})$
$$\exists \epsilon : |(f+g)(\overline{x}) - m'| > \epsilon$$
Q.E.D.

 A: The proof you trie to give does not work and you do not need to prove the non existence of the limit by using the definition.
Here it is,what you can do:

If the limit of $f+g$ at $a$ existed then $$\lim_{x \to a}g(x)=\lim_{x\to a}(f+g)(x)-\lim_{x\to a}f(x)$$would exist by algebra of limits.

A: You can write that $g$ has no limit in $a$ as:
$\exists \varepsilon>0\mid \forall \delta>0,\ \exists (x,y)\in(a-\delta,a+\delta)^2\text{ st. } |g(x)-g(y)|\ge \varepsilon\tag{1}\label{eq1}$
Since $f$ has a limit $\ell$ in $a$ we can choose the same epsilon and write:
$\exists \delta_1>0 \mid \forall x\in(a-\delta_1,a+\delta_1),\ |f(x)-\ell|<\frac{\varepsilon}4\tag{2}\label{eq2}$
Now we can combine both while letting $\delta<\delta_1$ so that the interval $(a-\delta,a+\delta)\subset(a-\delta_1,a+\delta_1)$ and any $x$ or $y$ verifying $\eqref{eq1}$ also verifies $\eqref{eq2}$.
Note: the restriction on $\delta$ is not important, we are more interested in small values than in big ones. The no limit fact is about however close $x,y$ may be we can always separate $g(x)$ and $g(y)$. We are not so much interested about $x,y$ being far apart.

Thus under the conditions of $\eqref{eq1}$ we have:
$\bigg|(f+g)(x)-(f+g)(y)\bigg|=\bigg|f(x)-f(y)+g(x)-g(y)\bigg|\ge\bigg||g(x)-g(y)|-|f(x)-f(y)|\bigg|$
And since $\begin{cases}|f(x)-f(y)|=|f(x)-\ell-f(y)+\ell|\le|f(x)-\ell|+|f(y)-\ell|\le \frac{\varepsilon}4+\frac{\varepsilon}4\le \frac{\varepsilon}2\\|g(x)-g(y)|\ge \varepsilon\end{cases}$
We get what we wanted to ensure:
$$\bigg|(f+g)(x)-(f+g)(y)\bigg|\ge \frac{\varepsilon}2\tag{3}\label{eq3}$$
This later $\eqref{eq3}$ is the pendant of $\eqref{eq1}$ for $(f+g)$ and we conclude that $f+g$ has no limit either.
