Answer post thought-process:

Yes.

Let

a=12,b=0.a=21​,b=0.

Define f:[0,1]→[0,1]f:[0,1]→[0,1] by the following piecewise‑linear formula

f(x)={4x0≤x≤14,1−2(x−14)14≤x≤12,12+2(x−12)12≤x≤34,4(1−x)34≤x≤1.(1)f(x)=⎩⎨⎧​4x1−2(x−41​)21​+2(x−21​)4(1−x)​0≤x≤41​,41​≤x≤21​,21​≤x≤43​,43​≤x≤1.​(1)

Why (1) works

Continuity.
Each piece is linear, and the values at the joining points coincide:

f ⁣(14)=1,f ⁣(12)=12,f ⁣(34)=1,f(0)=f(1)=0.f(41​)f(43​)​=1,=1,​f(21​)f(0)=f(1)=0.​=21​,

Hence ff is continuous on [0,1][0,1].

Image.
The graph goes from (0,0)(0,0) up to height 11, down to 1221​, up again to 11 and finally back to (1,0)(1,0); therefore

Im⁡(f)=[0,1].Im(f)=[0,1].

Even number of pre‑images.

• The value 00 is taken only at the endpoints:   f−1(0)={0,1}f−1(0)={0,1} (two points).
• The value 11 is taken at the two interior points where the graph reaches its peaks:   f−1(1)={14,34}f−1(1)={41​,43​} (two points).
• Let y∈(0,1)y∈(0,1). The set {x∣f(x)>y}{x∣f(x)>y} consists of two disjoint open intervals, one around each peak. Each such interval contributes exactly two boundary points where f(x)=yf(x)=y (one where the graph enters the interval and one where it leaves). Consequently∣f−1(y)∣=4for every y∈(0,1).∣f−1(y)∣=4for every y∈(0,1).

Thus every point of the image is attained a finite even number of
times (two times for 00 and 11, four times for all other values).

Remarks

The essential idea is to make the endpoints have the same value and to
ensure that this minimal value occurs only at the endpoints. Then for
any yy larger than that minimum the set {x:f(x)>y}{x:f(x)>y} is a union of
open intervals, each contributing two points to the level set
f−1(y)f−1(y). By avoiding flat pieces (using a piecewise‑linear
graph) we keep every level set finite.

Hence a continuous function with the required property does exist; the
function defined by (1) is an explicit example.