real analysis - Proof that a linear transformation is continuous

I got started recently on proofs about continuity and so on. So to start working with this on $n$-spaces I've selected to prove that every linear function $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous at every $a \in \mathbb{R}^n$. Since I'm just getting started with this kind of proof I just want to know if my proof is okay or if there's any inconsistency. My proof is as follows:

Since $f$ is linear, we know that there's some $k\in \mathbb{R}$ such that $|f(x)|\leq k|x|$ for every $x\in \mathbb{R}^n$, in that case let $a\in \mathbb{R}^n$ and let $\varepsilon >0$. Consider $\delta = \varepsilon /k$ and suppose $|x-a|<\delta$, in that case we have:

$$|f(x)-f(a)|=|f(x-a)|\leq k |x-a|

And since $|x-a|<\delta$ implies $|f(x)-f(a)|<\varepsilon$ we have that $f$ is continuous at $a \in \mathbb{R}^n$. Since $a$ was arbitrary, $f$ is continous in $\mathbb{R}^n$. Is this proof fine? Or there was something I've missed on the way?

Answer

This proof is correct modulo result you stated in the begining, i.e.
$$

\text{there exist $k\in\mathbb{R}$ such that $|f(x)|\leq k|x|$ for all $x\in\mathbb{R}^n$}
$$
Proof of this fact is much more interesting and uses compactness of unit ball in $\mathbb{R}^n$.