Given f : R → R is continuous on [a,b] and sgn( f(a) )sgn( f(b) )= -1

Show that ∃ x∈[a,b] : f(x) = 0

Looks to me like a variation of Rolle's theorem.

No, I would restate it as "Many [relatively] easy statements in mathematics are [relatively] hard to prove."

Two cases: either

f(a) < 0, f(b) > 0

or

f(a) > 0, f(b) < 0

Given f(x) is continuous from a to b, it must not jump. And yet it must cross 0 in order to go from <0 to >0 (or the other way around). Therefore it must at some point equal 0.

Is this proof valid?

I would like to get a more Rigour proof

i'm not the best at writing proofs, sorry

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Looks to me like a variation of Rolle's theorem.

No, I would restate it as "Many [relatively] easy statements in mathematics are [relatively] hard to prove."

Two cases: either

f(a) < 0, f(b) > 0

or

f(a) > 0, f(b) < 0

Given f(x) is continuous from a to b, it must not jump. And yet it must cross 0 in order to go from <0 to >0 (or the other way around). Therefore it must at some point equal 0.

Is this proof valid?

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What are the advantages and disadvantages of the PDF format. I am sure they will make your work easier than before. Highly recommended to you guys!