Proof that if a sequence of random variables converges weakly to a constant, then it converges to it in probability Is my proof correct? A sequence of random variables {$\xi_n$}$\xrightarrow{w}c$ means by definition that $F_{\xi_n}(t)\rightarrow F_c(t)$ for every $t$ such that $F_c(t)$ is continuous. 
So, i have $F_{\xi_n}(t)-F_c(t)\rightarrow0$. $(1)$
$F_c(t)$ is not continuous at $t=c$, so we need to consider $t>c$ and $t<c$. 
For $t>c$, $F_c(t)=1$, so $P(\xi_n\leq t)-1\rightarrow 0$
For $t<c$, $F_c(t)=0$, so $P(\xi_n\leq t)-0\rightarrow 0$
Now, i substitute $t=\varepsilon+c$ and get:
For $\varepsilon>0$, $P(\xi_n-c\leq \varepsilon)\rightarrow 1$ (3)
For $\varepsilon<0$, $P(\xi_n-c\leq \varepsilon)\rightarrow 0$ $\implies$ $P(\xi_n-c> \varepsilon)\rightarrow 1$ $\implies$ (?) $P(\xi_n-c\geq \varepsilon)\rightarrow 1$ (4)
Now, i say that ($(3)$ and $(4)$) is equal to $P(|\xi_n-c|\leq \varepsilon)\rightarrow 1$ for $\varepsilon>0$, which is equal to $P(|\xi_n-c|> \varepsilon)\rightarrow 0$, which is convergence in probability by definition. But for that, i need implication "(?)". Do i have it? 

Well, implication (?) is not true on its own in general - it would fail if many of the $\xi_n - c$ were equal to $\varepsilon$ with positive probability - but the conclusion is in fact true in this case.  A quick fix is just to use $\varepsilon/2$ instead of $\varepsilon$.  
Notation comment: people usually expect $\varepsilon$ to be a positive quantity, so if you want a negative quantity use $-\varepsilon$ instead. 
Now for any $\varepsilon >0$, you can say $P(\xi_n - c \le -\varepsilon/2) \to 0$, meaning that $P(\xi_n - c > -\varepsilon/2) \to 1$.  But any quantity $> -\varepsilon/2$ is certainly $\ge -\varepsilon$, so you have $P(\xi_n - c \ge -\varepsilon) \ge P(\xi_n - c > -\varepsilon/2)$.  By the squeeze theorem, you conclude  $P(\xi_n - c \ge -\varepsilon)\to 1$.
