If $f$ is continuous then there exists $x\in [0,1]: f(x)=x$ I wish to prove the following by contradiction:

Let $f:[0,1]\rightarrow[0,1]$ be a continuous function. Prove that there exists $x\in [0,1]$ such that $f(x)=x$.

Proving this directly, one would have to define a new function $g(x):=f(x)-x$ and prove that there exists $x\in [0,1]$ such that $g(x)=0$ using Intermediate Value Theorem, hence $f(x)=x$. But I'm struggling to elaborate a proof by contradiction.
My approach is as follows. Suppose by contradiction that it does not exist $x\in [0,1]$ such that $f(x)=x$. Then for all $x\in [0,1]$, $f(x)\neq x$. Then $f(x)>x$ or $f(x)<x$ for all $x$. This is where I'm not sure if it's correct.
If $f(x)>x$ for all $x\in [0,1]$, let $x:=1$. Then $f(1)>1 \implies f(1)\notin [0,1]$ which is a contradiction because $f$ was defined in $[0,1]$
If $f(x)<x$ for all $x\in [0,1]$, let $x:=0$. Then $f(0)<0 \implies f(0)\notin [0,1]$ leading to the same contradiction as above.
I'm skeptical because my contradiction does not involve $f$s continuity. Is the proof correct? If not, how can I approach this by contradiction? Thanks.
 A: For each individual $x$, we can conclude that $f(x) > x$ or $f(x) < x$; this doesn't imply that the same inequality holds for all $x \in [0, 1]$ - this is exactly where we have to use the fact that $f$ is continuous. As an example of this can fail without using the continuity of $f$, consider the function which is $1$ on $[0, 1/2]$ and $0$ otherwise.
Using the continuity of $f$ to prove that the inequality is the same for all points in $[0, 1]$ is really equivalent to the argument using the IVT that you outlined at the beginning of your question: Since $f(x) - x$ has no zeros, it can't change sign, implying what we want.
A: If you insist on a proof by contradiction, here is one, based on your argument:
Suppose $g$ has no zero. Then, by the Intermediate Value Theorem, $g(x)>0$ for all $x\in [0,1]$ or $g(x)<0$ for all $x\in [0,1]$. (Here you have used the continuity of $g$, which is equivalent to the continuity of $f$.) In the first case, $g(1)>1$. In the second case, $g(0)<0$. In either case, you get contradiction.
The standard proof is a direct proof, not by contradiction, and I find it simpler:
If $f(0)=0$ or $f(1)=1$, then you have found a fixed point of $f$.
Otherwise, $f(0)>0$ and $f(1)<1$ and so $g(0)>0$ and $g(1)<0$.
The Intermediate Value Theorem now gives you a zero of $g$ and this is a fixed point of $f$.
A: You could also word it like this: suppose that $f$ has no fixed point then $f(0) >0$ and $f(1) < 1.$ Hence $g(x) = x-f(x)$ is negative at $x =0$ and positive at $x =1.$ Since $g(x)$ is continuous, we must have $g(x) = 0$ for some $x \in (0,1)$ by the intermediate value theorem,
so $f(x) = x$ for this $x,$ contrary to our assumption. 
