denote the sequence of distribution $\mu_n = \delta_{1/n}(dx)$. I want to see that this sequence converges weakly to $\delta_0(dx)$, where $\delta_x(A)= 1_{A}(x)$ (i.e. that its distribution functions $F_n(y) \rightarrow F(y)$ for all points of continuity of $F(y) = \mu((-\infty,y])$).
First of all, what is the meaning of dx here, and second, why is $F_n (y) = 1\{{y \geq 1/n}\}$ and not $F_n (y) = 1\{{-1/n \leq y \leq 1/n}\}$?
Don't bother too much about the $dx$ notation. In my view it is inconsistent here and personally I would just say that $\mu_n=\delta_{\frac1n}$ (or at most $\mu_n(dx)=\delta_{\frac1n}(dx)$).
In general $\delta_x(A)=1_A(x)$ where $1_A$ denotes the indicator function of set $A$.
Applying that here we find
$F_n(y)=\delta_{\frac1{n}}((-\infty,y])=1_{(-\infty,y]}(\frac1{n})$.
Observe that $F_n$ takes values in $\{0,1\}$ with $F_n(y)=1\iff y\geq\frac1{n}$.
This allows us to write $F_n=1_{[\frac1{n},\infty)}$ or with argument $F_n(y)=1_{[\frac1{n},\infty)}(y)$
Likewise we find that $F=1_{[0,\infty)}$ and discern that $F$ has $\mathbb R-\{0\}$ as set of continuity points.
It remains to be shown now that $\lim_{n\to\infty}1_{[\frac1{n},\infty)}(y)=1_{[0,\infty)}(y)$ for every $y\in\mathbb R-\{0\}$.
Give it a try yourself.