Please verify my proof for a continuous function attaining a minimum value on an interval. I was given the following problem in my exam:

A function continuous on $[a,b]$ attains a minimum value on $[a,b]$.

Note: proof should not involve compact sets or sequences.


My proof:

Using the completeness axiom in Real numbers:

For any closed interval in $\mathbb R $ ,there exists a greatest lower
    bound, called the infimum, which is equal to the minimum of that set.
Since the function is continuous and the domain is a closed interval,
    it follows that its range is also a closed interval, so the completeness axiom holds for the Range of that function,
    i.e. We get a minimum. 
ADD:To see why the range would be a closed set,
    Consider $c$ to be a point in the interval $[a,b]$,
    and assume it is one of the end points of the function's range. Then if the range was not a closed interval, then $f(c)$ would not be in the range,and thus :$$ \lim_{x\to c} f(x) = f(c) $$ would not hold.
    Since the function would not be defined on the point $c$, 
    and we would get a contradiction that the function is continuous
     on the interval $[a,b]$. Therefore it is a closed set.


Is it considered a valid proof? Please send your suggestions. Thanks.

This sentence is fluff:
"For any closed interval in R ,there exists a greatest lower bound, called the infimum, which is equal to the minimum of that set."
The infimum exists regardless of whether the function attains its infimum
That the image of any continuous function over a closed domain is closed is an interesting implication of continuity.  If this has been proven in class, then you can use it.  If not, then you must prove it.
It is a corollary to the fact that for any continuous function, for every open set in the image the pre-image is an open set.
But, once you have that the image is a closed set, then the set contains its infimum.
