the question was asked before but I didn't get the hints or the attempted answer in: Continuity of a constant function
Edited: I applied what was mentioned on the link I eneded up with : `
l f(x)-c1l<= lf(x)l +lc1l <= lc1l +l-c2l simplifying stuff I get : lf(x)-c1l<= lf(x)l -lc2l <=0
`
So I conluded from this that lf(x)-c1l<=0 hence f(x)=c1... f constant I am not sure that doing this right since I ignored the term lf(x)l -lc2l ??
You don't need to use the triangle inequality unless you know what you are doing, because you may lose too much information.
Here, suppose that $f(x_0)=c_1$ for some $x$. If you choose $\varepsilon$ as suggested in the linked question, you get from continuity that there exists $\delta>0$ such that on the whole interval $(x_0-\delta,x_0+\delta)$, you have $$\tag1 |f(x)-f(x_0)|=|f(x)-c_1|<\varepsilon<|c_j-c_1|,\ \ \text{ for all } j. $$ But $f$ takes finitely many values, so $f(x)=c_j$ for some $j$. That makes the inequality in $(1)$ impossible unless $f(x)=c_1$. Thus $f(x)=c_1$ for all $x\in (x_0-\delta,x_0+\delta)$.