Show the estimate of the distribution function and show the convergence in distribution

Let $$X_n$$ be a family of random variables an $$X$$ a rv with $$X_n \overset{D}\rightarrow X$$. Let $$a_n$$ a sequence of real numbers with $$a_n \overset{n \rightarrow \infty} \rightarrow a$$. We say that the Distribution function of $$X$$ is continuous.

a) Proof that for all $$x \in \mathbb{R}$$ an every $$\epsilon >0$$ there exist a $$N_{\epsilon,x} \in \mathbb{N}$$, thus for all $$n \geq N_{\epsilon,x}:$$

$$P(X+a \leq x - \epsilon) - \epsilon \leq P(X_n + a_n \leq x) \leq P(X+a \leq x + \epsilon) + \epsilon$$

b) Show that

$$X_n + a_n \overset{D}\rightarrow X+a$$ for $$n \rightarrow \infty$$

Can anybody give me a hint? I have no idea how to start with the excersice.

$$P(X_n+a_n \leq x) \leq P(X_n \leq x-a+\epsilon)$$ for $$n$$ sufficiently large. Now use the fact that $$\lim \sup P(X_n \leq y) \leq P(X\leq y)$$ for all $$y$$. Hence $$\lim \sup P(X_n+a_n \leq x-a+\epsilon) \leq P(X \leq x-a+\epsilon)$$. It follows that $$P(X_n+a_n \leq x) < P(X+a \leq x+\epsilon)+\epsilon$$ for $$n$$ sufficiently large. The left hand inequality in a) is proved in a similar manner.
b) follows immeditely from a): if $$x$$ is a continuity point for $$X+a$$ then the right and left extremes in the inequality in a) both tend to $$P(X+a \leq x)$$.