$X'$ finite-dimensional implies $X$ finite-dimensional

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 $\dim(X) \leq \dim(X'') < \infty.$

It strikes me that the answer provided by our colleague Eddie is quite on the mark; the solution to this problem seems to center on two essential facts: first, that for finite dimensional $Y$, $Y'$ is also finite dimensional; second, the the natural mapping from $Y$ to its double dual $Y''$ given by evaluation on elements of $Y'$ is injective; that is, the functional $\hat y \in Y''$ defined by

$\hat y(\phi) = \phi(y), \; \forall \phi \in Y', \tag 0$

is determined solely by $y$; there is no other $z \in Y$ such that

$\hat z(\phi) = \phi(y), \; \forall \phi \in Y'; \tag{0.1}$

these two principles combine to yield the desired result, as Eddie has indicated, and as I discuss at some greater length below.

What I have done in preparing this answer is to try and expand upon these two principles and work out how they operate jointly and severally in some detail, as much for the sake of my own recollection as anything. So this answer may be a tad on the long-winded side; I thank you all in advance for your patience.

Now let' see . . .

For any finite dimensional normed space $Y$, $Y'$ is also finite dimensional, with

$\dim Y' = \dim Y; \tag 1$

for let

$y_1, y_2, \ldots, y_n \in Y \tag 2$

be a basis; then set

$\phi_i \in Y', \; \phi_i(y_j) = \delta_{ij}, \; 1 \le i \le n; \tag 3$

for $y \in Y$ we may write

$y = \displaystyle \sum_1^n \alpha_i y_i; \tag 4$

thus for $\phi \in Y'$,

$\phi(y) = \displaystyle \sum_1^n \alpha_i \phi(y_i); \tag 5$

I claim

$\phi = \displaystyle \sum_1^n \phi(y_j) \phi_j; \tag 6$

for with $y$ as in (4),

$\left (\displaystyle \sum_1^n \phi(y_j) \phi_j \right )(y) = \left (\displaystyle \sum_1^n \phi(y_j) \phi_j \right ) \left ( \displaystyle \sum_1^n \alpha_i y_i \right ) = \displaystyle \sum_1^n \phi(y_j) \phi_j \left ( \displaystyle \sum_1^n \alpha_i y_i \right )$ $= \displaystyle \sum_{i, j = 1}^n \phi(y_j) \alpha_i \phi_j(y_i) = \sum_{i, j = 1}^n \phi(y_j) \alpha_i \delta_{ij} = \sum_1^n \alpha_i \phi(y_i) = \phi(y), \tag 7$

showing the $\phi_i$ span $Y'$; the $\phi_i$ are also linearly independent, for if

$\displaystyle \sum_1^n \beta_j \phi_j = 0, \tag 8$

with say $\beta_k \ne 0$ we have

$\beta_k \phi_k = -\displaystyle \sum_{i = 1, i \ne k}^n \beta_i \phi_i; \tag 9$

now if we apply this to $y_k$ we find

$\beta_k = 0, \tag{10}$

a contradiction. So the $\phi_i$ are indeed linearly independent, and as such form a basis for $Y'$; thus, (1) holds.

Next, replacing $Y$ with $Y'$ we see that

$\dim Y'' = \dim Y'; \tag{11}$

now consider the map

$Y \to Y'', \; y \to \hat y, \; \hat y(\phi) = \phi(y), \; \forall \phi \in Y'; \tag{12}$

I claim $y \to \hat y$ is injective; if

$\exists y \in Y, \; \hat y(\phi) = 0, \; \forall \phi \in Y', \tag{13}$

then

$\phi(y) = \hat y(\phi) = 0, \forall \phi \in Y'; \tag{14}$

so we need to show that

$\phi(y) = 0, \; \forall \phi \in Y' \Longrightarrow y = 0; \tag{15}$

this can be accomplished via the Hahn-Banach theorem; if $y \ne 0$ we my consider the subspace

$\{0\} \ne \{\alpha y \} \subset Y; \tag{16}$

define $\phi$ on this subspace by

$\phi(\alpha y) = \alpha \vert y \vert; \tag{17}$

then

$\vert \phi(\alpha y) \vert = \vert \alpha \vert y \vert \vert = \vert \alpha \vert \vert y \vert, \tag{18}$

which shows that $\phi$ is bounded on $\{\alpha y \}$ with bound $\vert y \vert$ (note $\vert \phi y \vert = \vert y \vert$); thus $\phi$ may be extended to $Y$ with the same bound; but then we have $\phi \in Y'$ with $\phi(y) \ne 0$; this contradiction proves (15). Thus, $y \to \hat y$ is injective.

Therefore,

$\dim Y \le \dim Y'' = \dim Y'. \tag{19}$