\begin{array}{l}{\text { Let }(X,\|\cdot\|) \text { be a normed linear space and let }\left\{x_{n}\right\} \text { be a sequence in }} \\ {X \text { with } x=\lim _{n \rightarrow \infty} x_{n} . \text { Assume that }\left\|x_{n}-y\right\| \leq a \text { for all } n \in \mathbb{N} . \text { Show that }\|x-y\| \leq a \text { . }}\end{array}
I have the sense that all $x_n$ is in the closed ball centered at $y$ hence its limit should also be included in this closed ball with radius $a$. How to write the proof formally? Any help is appreciated.
The usual trick is to use the inequality $$||x - y|| \le ||x - x_n|| + ||x_n - y||.$$
This follows from the triangle inequality. The point is that you know something about $||x - x_n||$ and something about $||x_n - y||$.
Another trick is that $u \le v$ is equivalent to $ u \le v + \varepsilon$ for every $\varepsilon > 0$. You may want to try proving this.