Convergence, Banach-Steinhaus Let $(\phi_n)$ be a sequence of continuous linear functionals from $X$ - Banach space to $\mathbb{R}$ such that $(\phi_n(x))$ is convergent for all $x\in X$. Show that if a sequence $(x_n)$ is convergent in X norm, then $(\phi_n(x_n))$ is convergent.
My attempt: 
All assumptions of Banach-Steinhaus are satisfied. So we have $\sup_n\|\phi_n\|\le M, \:\:M>0$.
So we have $|\phi_n (x_n)|\le M\|x_n\|$ for all $n$. If $(x_n)$ is convergent, so is $(\phi_n (x_n))$.
Is this ok?
Edit: Let $x\in X$ be a limit of $(x_n)$ and $\phi(x)$ be a limit of $(\phi_n(x))$. 
$|\phi_n (x_n-x)|=|\phi_n (x_n)-\phi_n(x)|\le M\|x_n-x\| \rightarrow 0 \:\:\:\:$ letting $n$ go to $\infty$

Concerning your last edit, you're almost there. You don't have to show anything is a Cauchy sequence when you already know the limit:
$$
\begin{align}
      |\phi_{n}(x_{n})-\lim_{n}\phi_{n}(x)| & \le |\phi_{n}(x_{n}-x)|+|\phi_{n}(x)-\lim_{n}\phi_{n}(x)| \\
         & \le M\|x_{n}-x\|+|\phi_{n}(x)-\lim_{n}\phi_{n}(x)|
\end{align}
$$
Each term on the right tends to $0$ as $n\rightarrow 0$. A few simple details and you're there.
