More specifically:
Given the space of continuous functions $\mathbb{R}\rightarrow\mathbb{R}$ where convergence is equivalent to uniform convergence on compacta (i.e., compact sets), prove that there is no norm for which convergence in norm is equivalent to simple convergence in the space.
My main difficulty is that convergence in norm is so abstract -- I can assign norms to functions all but arbitrarily as long as I follow the axioms. One example is that the function $$f_n(x) = \begin{cases} 0, &x < n \\ x-n, & x\geq n \end{cases}$$ converges to zero on compacta (just take an interval $[-k, k]$), it "obviously" is not getting close to zero, and yet I can't seem to derive a contradiction. After all, what's to stop $$\|f_n\|\rightarrow 0$$ since the axioms for norm only give information about when the norm equals zero?
The given space is a locally convex space (LCS). The topology is generated by the seminorms $p_K$, where $K$ is compact and $p_K(f) = \sup_{t\in K}|f(t)|$. By Kolmogorov, a LCS is normable if and only if there exists a bounded neighborhood of zero. Hereby, a set $S$ is called bounded if for each neighborhood $U$ of zero there exists $k > 0$ such that $S\subset kU$. You can use this criterion to prove that the present LCS is not normable.