Elementary Approach using Bolzano's Theorem and not IVT :
Let $g(x) = f(x) - f(x-1)$. This function is continuous in $[0,2]$ as an operation of continuous functions.
Now, $g(1) = f(1)- f(0)$ and $g(2) = f(2)-f(1)$.
$$g(1)\cdot g(2) =f(1)f(2)-f(1)^2 - f(2)f(0)+ f(0)f(1)=2f(0)f(1) - f(1)^2-f(0)^2 $$
$$g(1)\cdot g(2) = -\big[f(1)-f(0)\big]^2 \leq 0 $$
Thus, there exists $a \in [0,2]$ such that $g(a) =0$ by Bolzano's Theorem.
Functional Analysis (sorry for this) :
Essentialy, we want to show that the equation
$$f(x) - f(x-1) =0$$
has a solution.
Let $T$ be a linear operator $T:C[0,2] \to C[0,2]$ such that $Tf(x) = -f(x-1)$. One can easily see that $T$ is a bounded linear operator $\in B\big(C[0,2]\big)$. Also, one can take $0$ to be a function defined in $C[0,2]$. Then, the equation is re-written as :
$$f + Tf = 0$$
Note that $C[0,2]$ is a Banach Space and since $T \in B\big(C[0,2]\big)$ then there exists a unique solution in $C[0,2]$, meaning that there exists such function $f(x) \in C[0,2]$ which of course implies that the equation has a solution for $x \in [0,2]$.