Solution: To show that $\mathbb{R} $ is not algebraically closed, we need to find a polynomial in $\mathbb{R} [x]$ which does not have a zero in $\mathbb{R} $. The polynomial $p(x) = x^2 + 1$ does not have a zero in $\mathbb{R} [x]$ and hence, $\mathbb{R} $ is not algebraically closed.

Let $\mathbb{F} $ be an algebraically closed field. We need to show that an $\mathbb{F} $ is infinite. Let us assume that $\mathbb{F} $ is a finite field with $a_1,a_2,\dots,a_n$ elements. Consider the polynomial \[ p(x) = \prod_{i=1}^n \left( x - a_i \right) + 1. \] Since the coefficient of $p(x)$ lies in $\mathbb{F} [x]$, it is clear that $p(x) \in \mathbb{F} [x]$. But note that for each $i= 1,2,\dots,n$ \[ p\left( a_i \right) = 1 \neq 0. \] Thus, $p(x)$ does not have a root in $\mathbb{F} [x]$, which gives a contradiction to the fact that $\mathbb{F} $ is an algebraically closed field. Thus, $\mathbb{F} $ must be an infinite field.