$[F(a):F]<\infty \implies a$ algebraic over F Let $E/F$ be an extention field and $a\in E$. We want to show that 
$$[F(a):F]<\infty \implies a\text{ algebraic over } F$$
without the theorem which tells us that every finite extension is algebraic.
Proof.  Let $[F(a):F]<\infty$. If $a\in E$ was transcendental over $F$, then $F(a)\cong F(x)$. But, we know that $[F(x):F]=\infty$. So, $[F(a):F]=\infty$, contradiction.
Is this proof correct?
Thank you

If you know that $\langle a \rangle$ spans $F(a)$ (as an $F$-vector space):
Observe that $F(a)$ is singly generated over $F$.  In particular, $\{1, a, a^2, \dots\}$ is a spanning set of $F(a)$ (seen as an $F$-vector space).  Since $[F(a):F] = n < \infty$, there are $f_0, \dots, f_n \in F$ such that $$\sum_{i=0}^n f_i a^i = 0  \text{.}$$  I.e., $\{1, a, a^2, \dots, a^n\}$ is $F$-linearly dependent.  But this says $a$ is a root of a degree $n$ polynomial with coefficients in $F$, so $a$ is algebraic over $F$.
If you do not know that $\langle a \rangle$ spans $F(a)$ (as an $F$-vector space):
Since $[F(a):F] = n < \infty$, $F(a)$ is an $n$-dimensional vector space over $F$.  Then $\{1, a, a^2, \dots, a^n\}$ is a list of $n+1$ elements of this vector space, so is $F$-linearly dependent.  I.e., there are $f_0, \dots, f_n \in F$ such that $$\sum_{i=0}^n f_i a^i = 0  \text{.}$$  But this says $a$ is a root of a degree $n$ polynomial with coefficients in $F$, so $a$ is algebraic over $F$.
