Suppose that $E$ is finite dimensional. Let $x_1, \ldots, x_n$ be a basis for $E$. Consider the coordinate functionals $f_1, \ldots, f_n$ given by
$$\langle f_i, \sum_{k=1}^n \lambda_k x_k \rangle = \lambda_i\quad (i\leqslant n)$$
Prove that $f_1, \ldots, f_n$ are linearly independent. Having this done, note that they span $E^\prime$. Indeed, if $f$ is any functional on $E$, then
$$\langle f, \sum_{k=1}^n \lambda_k x_k\rangle = \sum_{i=1}^n \lambda_k\langle f,x_k\rangle $$
so $f = \sum_{i=1}^n \langle f, x_k \rangle f_i$ because
$$\langle \sum_{i=1}^n \langle f, x_k \rangle f_i, \sum_{k=1}^n \lambda_k x_k\rangle = \sum_{k=1}^n\sum_{i=1}^n \lambda_k \langle f_i, x_k\rangle \langle f, x_k \rangle $$
Note that $\langle f_i, x_k\rangle = 0$ unless $i=k$ in which case this is equal to 1. Consequently, the right hand side is equal to $\sum_{i=1}^n \lambda_k\langle f,x_k\rangle$.
We thus proved that $f_1, \ldots, f_n$ form a basis for $E^\prime$. So $E$ and $E^\prime$ have the same dimension.
Yes. Your argument is valid. In a nutshell, your argument is this:
Take any $f \in X^*$. Since $X$ is finite dimensional, $f$ is bounded. So, $f \in X'$. So, $X^* \subseteq X'$, as desired.
When $X$ is infinite dimensional, we can always construct an unbounded linear function $f$ on $X$. Finding an explicit construction of such an $X$ is tricky. However, it suffices to find an example that works on a vector space of countably infinite dimension (it is non-trivial to see that this is indeed enough; you'll need to inject the axiom of choice appropriately).
To that end: let $\{e_k\}_{k \in \Bbb N}$ be a (Hamel-)basis for $X$ consisting of unit-vectors. We can the define a linear functional by
$$
f(e_n) = n
$$
Hopefully you can see that this must be unbounded.
Best Answer
Since $X'$ is finite dimensional we can choose a basis $x_1', \cdots, x_n'$ of $X'$. The vectors $(x_i'')_{i \in \{1, \cdots, n \}}$ defined by $x_i''(x_j') = \delta_{ij}$ are a basis of $X''$. This means $X''$ is finite dimensional too. The canonical inclusion $J: X \to X''$ defined by $$Jx: X' \to \mathbb{K}, \ (Jx)(x') := x'(x)$$ is injective we get $\dim(X) \leq \dim(X'') < \infty.$