Proof of Weak Convergence Suppose that we have two cumulative distributions $F_{n}(x)=\left\{\begin{matrix}
1, \ x\in [\frac{1}{n},\infty) \\  
0, \ otherwise
\end{matrix}\right.$
and $F(x) =\left\{\begin{matrix}
1, \ x\in [0,\infty) \\  
0, \ otherwise
\end{matrix}\right.$
In order to prove that $F_{n}(x)$ converges weakly to $F(x)$ I have to prove that $lim_{n\infty}F_{n}(x) = F(x)$ for all $x \in \mathbb{R}$.

What I tried is the following, prove that $F_{n}(x)$ converges to $F(x)$ and let this convergence to fail for $x$ in a set of Lebesgue measure $0$.
That set that convergence fails I believe is $(0,\frac{1}{n})$, which has Lebesgue measure $\lambda(0,\frac{1}{n})=\frac{1}{n}\rightarrow 0$.
Eventually, $F_{n}(x)$ \converges to $F(x)$ for all $x\in \mathbb{R}$ except from a zero Lebesgue measure set.
Is the proof rigorous, or do I have to add more details to change something, and of course is it correct?

You're missing a small but key point of the definition of weak convergence: $F_n \stackrel{\text w}\to F$ if $\lim_{n\to \infty} F_n(x) = F(x)$ for all $x$ that are continuity points of $F$. This is important because $x\mapsto \lim_n F_n(x)$ is not guaranteed to be a valid CDF, and your example shows this because $\lim_n F_n(0) = 0$ and this would violate the right-continuity requirement of a CDF. So in this case we just need to show that the limit holds for all $x \neq 0$ and that is straightforward.
The Lebesgue measure of the set of disagreements is also not important, because if two CDFs disagree on even a single point then the corresponding probability measures are different and we've got two distinct distributions. A continuous random variable can be modified on a set of measure zero without changing anything, but modifying a CDF on a set of Lebesgue measure zero is different. That's because we're actually changing the measure of a collection of sets when we change the value of a CDF. For example, if I have a CDF $G$ and I change $G(1)$ from $0.6$ to $0.65$, say, I've actually changed the measure of the set $(-\infty, 1]$ and that changes the distribution, despite $\{1\}$ being a Lebesgue null set.
