We need to prove the following proposition:

Let $F$ be a field containing $\mathbb{R}$ with the property that $\dim_{\mathbb R}F < \infty$. Then either $F \cong \mathbb R$ or $F \cong \mathbb C$.

We give three proof in the following, where the first one is the most simple, and the last one is the most complicated.

Proof 1. By the uniqueness of the algebraic closure, we have an embedding $F \hookrightarrow \mathbb C$, hence we have $\mathbb R \subset F \subset \mathbb C$. The result follows from $[\mathbb C:\mathbb R]=2$, because this excludes the existence of proper intermediate fields.

Proof 2. Since $F$ is finite dimensional over $\mathbb{R}$ it is algebraic over $\mathbb{R}$. This is a basic fact about field extensions: if $a\in F$, then $1,a,a^2,\dots,a^n$ are not linearly independent over $\mathbb{R}$, where $n=\dim_{\mathbb{R}}F$. So every element of $F$ is the root of a polynomial with coefficients in $\mathbb{R}$.

If $-1$ is not a square in $F$, we can add a square root $j$ (with $j^2=-1$), so we have $F[j]$ which is finite dimensional over $F$, hence also over $\mathbb{R}$ by the dimension formula. Now $\mathbb{R}[j]$ is isomorphic to $\mathbb{C}$ and we have the chain $\mathbb{R}\subseteq \mathbb{R}[j]\subseteq F[j]$ which tells us that $F[j]=\mathbb{R}[j]$, because $\mathbb{C}$ is algebraically closed. Thus $F=\mathbb{R}$.

Otherwise $-1$ is a square in $F$, so $F$ is algebraic over $\mathbb{R}[j]$ (where $j^2=-1$) which is algebraically closed, being isomorphic to $\mathbb{C}$. Hence $F=\mathbb{R}[j]$ and so $F\cong\mathbb{C}$.

Proof 3. Let $x\in F\setminus\mathbb{R}1$. Since $F$ is finite-dimensional, say of dimension $N\geq1$, the family of $N+1$ vectors $(1,x,x^2,\ldots,x^N)$ is a dependent family. Hence there exists some real numbers $(\alpha_0,\ldots,\alpha_N)\neq(0,\ldots,0)$ such that $(\alpha_0,\ldots,\alpha_N)\neq(0,\ldots,0)$ and such that $$\alpha_N x^N+\cdots+\alpha_1x+\alpha_01=0.$$

Consider now the associated polynomial $P\in\mathbb{R}[X]$:$$P=\alpha_N X^N+\cdots+\alpha_1X+\alpha_0$$

Now, this polynomial can be factored as a product of irreducible polynomials in $\mathbb{R}[X]$; and by the Fundamental Theorem of Algebra we know that irreducible polynomials of $\mathbb{R}[X]$ are of degree $1$ or $2$. Without loss of generality we may hence assume that $P$ has the form:$$P=X^2+aX+b$$

(We can discard the case were the degree of $P$ is $1$, as this would mean that $x\in\mathbb{R}1$.)

Since $P$ is irreducible, we have $a^2/4-b>0$. Now, define the element $j$ of $F$ as $$j=\frac1{\sqrt{a^2/4-b}}x+\frac a{2\sqrt{a^2/4-b}}1.$$

We have: $$j^2=\frac1{a^2/4-b}\left(x^2+ax+\frac{a^2}41\right)=\frac1{a^2/4-b}\left(-b+\frac{a^2}4\right)1=-1$$ since $P(x)=x^2+ax+b1=0$.

Observe that there are exactly two elements of $F$ that have a square equal to $−1$, namely $j$ and $-j$. Indeed, if $u\in F$, then the following factorization holds true: $$u^2+1=(u+j)(u-j)$$

since $F$ is commutative, and since $F$ is an integral domain (since $F$ is a field), $$u^2+1=0\iff u=-j\ \text{or}\ u=j.$$

We now conclude by showing that $F=\operatorname{Span}\{1,j\}=\mathbb{R}1\oplus\mathbb{R}j$. First observe that the family $(1,j)$ is independent since $j\not\in\mathbb{R}1$ since $j^2=-1$. Now, the previous discussion shows that given an element $y\in F\setminus\mathbb{R}1$, there exists an irreducible polynomial $Q\in\mathbb{R}[X]$ of degree $2$, say $Q=X^2+\alpha X+\beta$, such that $Q(y)=0$, and the computation we performed earlier shows that the element $$u=\frac1{\sqrt{\alpha^2/4-b}}y+\frac\alpha{2\sqrt{\alpha^2/4-b}}1\in F$$

has a square equal to $−1$, hence this element is either $j$ or $-j$. Hence $$y=-\frac\alpha2\,1\pm\sqrt{\alpha^2/4-b}\,j$$ and in any case belongs to $\operatorname{Span}\{1,j\}$. Hence $F\subset\operatorname{Span}\{1,j\}$ hence $F=\operatorname{Span}\{1,j\}$. It is now obvious that $F$ is isomorphic (field isomorphism) to $\mathbb{C}$.