With the below, I'm a bit stuck as to where to proceed. Any tips would be greatly appreciated!
Proposition: If $x_{n} \rightarrow x_{0}$ implies $f(x_{n}) \rightarrow f(x_{0})$ then the transformation $f$ mapping $X$ to $Y$ is continuous at $x_{0} \in X$.
What I have so far:
If $x_{n} \rightarrow x_{0}$ implies $f(x_{n}) \rightarrow f(x_{0})$, then for all $\epsilon > 0$, there exists $\delta > 0$ such that $||f(x_{n}) - f(x_{0})|| < \epsilon$ for all points $x_{n} \in X$ for which $0 < ||x_{n} - x_{0}|| < \delta$.
The above looks like Rudin's definition for a limit. But given I've seen the above also used as the definition for continuity of $f$ at $x_{0} \in X$, is there anything else that needs to be done?
First of all, this is not special to normed linear spaces. It is true in any metric space. Secondly, the definition of continuity says for any $\epsilon >0$ there exist $\delta >0$ such that $\|f(x)-f(x_0)\| <\epsilon$ whenever $\|x-x_0\| <\delta$. This does not refer to any sequence $(x_n)$ converging to $x_0$.
To prove this result you have to prove by contradiction. All you have to do is write down the meaning if $f$ being not continuous at $x_0$. Can you do that?