This question was part of my assignment in Field Theory course and I was unable to solve it.
Let $E_1$ and $E_2$ be subfields of F and X a subset of F. If every element of $E_1$ is algebraic over $E_2$ , then prove that every element of $E_1(X) $ is algebraic over$E_2(X) $.
Attempt: it is not necessary that $E_1$ must be be proper subfield of $E_1$ .
Except the above statement which is due to comment of Gerry Myerson I don't have any other clues on how this question should be attempted.
So, please tell how to prove what was asked.
First, recall that if $L/K$ is a field extension then the set of elements of $L$ that are algebraic over $K$ is a subfield, $L^\mathrm{alg}$. see, for example this question.
In your situation, let $F^\mathrm{alg}$ be the set of elements of $F$ algebraic over $E_2(X)$. We want to show that $E_1(X)$ is a subfield of $F^\mathrm{alg}$. Certainly $E_1$ is a subset of $F^\mathrm{alg}$ since $E_1$ is algebraic over $E_2\subset E_2(F)$. Also, $X$ is clearly a subset of $F^\mathrm{alg}$ since it consists of elements of $E_2(X)$. Thus the field generated by $E_1$ and $X$, namely $E_1(X)$, must be a subfield of $F^\mathrm{alg}$, as needed.