Real function continuous on closed interval implies it is bounded - over-simple proof?? So here is my proof, which after looking up others seems to be too simple, or not rigourous enough, though I don't see why (hence I am asking!):
We take the contrapositive, and so prove that if $f$ is unbounded on $[a,b]$, then it is not continuous on this interval.
By hypothesis, $\exists c \in [a,b] : \displaystyle\lim_{x\to c} f(x) = \infty \implies f(c)$ is undefined. (or, $ \displaystyle\lim_{x\to c} f(x) \not= f(c)$ so we get discontinuity instantly)
Since $f(c)$ doesn't exist, the definition of continuity cannot be applied so $f$ must be discontinuous at at least $x=c$, as required.  


This addresses the proof proposed by the OP, showing that there exists some positive functions $f$ unbounded on $[0,1]$ such that the assertion $\lim\limits_{x\to c}f(x)=+\infty$ does not hold.

Choose some irrational $z$ in $[0,1]$ and define $f$ on $[0,1]$ by $f(x)=0$ if $x$ is irrational and $f(x)=1/|x-z|$ if $x$ is rational.
Then, $f$ is unbounded while, for every $c$ in $[0,1]$ there exists some $x$ as close to $c$ as one desires such that $f(x)=0$. Hence $\lim\limits_{x\to c}f(x)=+\infty$ does not hold.
