That's the statement I'm trying to prove. It's the converse of the result mentioned in this post: "Every linear mapping on a finite dimensional space is continuous".
It seems that a good way to start would be to replace 'continuous' with 'bounded' because they're the same thing for linear maps, and then attempt to show that a linear functional on an infinite-dimensional space can't be bounded. But that's where I get stuck.
We must speak in the context of a topological vector space in this question. Such a topology is not necessarily normable, therefore you see that you cannot talk about boundedness, orthonormality etc. in the ordinary sense.
You have also assumed that the basis of $V$ is countable, which it may not be. But these are small issues : you grasped the heart of the matter.
A linear functional, then, is continuous if and only if it is bounded. Bounded? Where is the norm on the space to speak of boundedness?
So, we have a notion of boundedness, that does not even mention the word "norm". A set $E$ in a topological vector space is bounded if given any other open set $V$ around $0$ (remember : open sets of the topological vector space ensure that addition and scalar multiplication are continuous), there is a real number $t$ large enough, such that the set $tV = \{tv : v \in E\}$ contains $E$. You can see why this reflects boundedness in the ordinary sense.
Now, continuity is equivalent to this : for any bounded set $E$, $l(E)$ should also be bounded i.e. bounded in the ordinary sense, as a subset of the real numbers.
To be clean on notation, let's define $l$ as follows : take a countable linearly independent set $e_i$, and extend it to a basis $e_i \cup f_\alpha$. Now define $l(e_i) = i$ and $l(f_{\alpha}) = 0$.
Why is $l$ not continuous now? Of course, it is easy to see, though I won't explain it, that the set $E = \{\sum a_{i}e_i + \sum a_\alpha f_{\alpha} : |a_i|,|a_{\alpha}| \leq 1 \forall i,\alpha\}$ is a bounded set : it's sort of like the open box in finite dimensions.
But $l(E)$ is unbounded, because $l(e_i)$ itself is unbounded!
That, is the proof of your assertion, with a lot of background added.