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.
 A: This is not valid. How do you know that the continuous image of a closed interval is a closed interval? This takes proof - indeed, it's essentially equivalent to what you're trying to prove in the first place.

You've now added an argument for why the continuous image of a closed set is closed. This has the right idea, but doesn't fully work. For example, "$\lim_{x\rightarrow c}f(x)=f(c)$" is incorrect: it should be $x\rightarrow d$ for some $d$ with $f(d)=c$, but you don't know such a $d$ exists!
It's better to argue as follows. Suppose $c$ is a point in the closure of $f([a, b])$; we want to show $c\in f([a, b])$. Since $c$ is in the closure of $f([a, b])$, we can find a sequence $d_i$ of points in $f([a, b])$ such that $d_i\rightarrow c$. 
Now, since each $d_i$ is in $f([a, b])$, we can find a sequence $e_i\in [a, b]$ such that $f(e_i)=d_i$.
Now what can you say about the sequence $e_i$?
A: Stating that the image of $f$ is a closed interval is essentially the same as proving $f$ attains a minimum value.
Indeed, if the image of $f$ is the interval $[c,d]$, then $c$ is the minimum value of $f$, because $c$ belongs to the image.
If you can prove that $f$ attains a minimum value $c$, the same proof applied to $-f$ shows $f$ attains a maximum value $d$. Together with the intermediate value theorem you see that the image of $f$ is $[c,d]$.

Here's a proof that doesn't (explicitly) use compactness and sequences.
First we prove that the image of $f$ has a lower bound.
Suppose not. Then on one of the intervals $[a,(a+b)/2]$ and $[(a+b)/2,b]$ the function has no lower bound. If it's the left interval call $a_1=a$ and $b_1=(a+b)/2$; otherwise set $a_1=(a+b)/2$ and $b_1=b$.
Now we can repeat the same argument obtaining a chain of intervals $[a_n,b_n]$ over none of which $f$ has a lower bound and
$$
[a,b]=[a_0,b_0]\supset[a_1,b_1]\supset[a_2,b_2]\supset\dotsb
$$
By completeness of the reals, there exists $r$ belonging to all those intervals. Since $f$ is continuous at $r$, there exist $l$ and $\delta>0$ such that, for $x\in(r-\delta,r+\delta)\cap[a,b]$, $f(x)>l$.
Choose $n$ such that $(b-a)/2^n<\delta$ and you get a contradiction, because $[a_n,b_n]\subset(r-\delta,r+\delta)$, and by construction the function has no lower bound on $[a_n,b_n]$, but $l$ is a lower bound.
Granted that the image of $f$ is lower bounded, call $c$ the greatest lower bound and suppose $c$ does not belong to the image of $f$.
Then the function
$$
g(x)=\frac{1}{f(x)-c}
$$
is continuous over $[a,b]$, but has no lower bound. Contradiction.
(There are a few points where the argument is only sketched, fill in the details.)
A: 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.
