I just read that as $\mathbb{N}$ is $C^*$-embedded in $\mathbb{R}$, then $\beta\mathbb{N}$ is embedded in $\beta\mathbb{R}$.
A subset $A$ of a space $X$ is said to be $C^*$-embedded if every continuous bounded $f:A\to \mathbb{R}$ has a continuous extension to the whole $X$. So, clearly by Tietze's theorem $\mathbb{N}$ is $C^*$-embedded in $\mathbb{R}$.
I don't see why this implies $\beta\mathbb{N}$ is embedded in $\beta\mathbb{R}$.
How can we construct this embedding?
This proof is based on Grumpy's suggestion. Let $A$ and $B$ the collection of all continuous function from $\Bbb{N}$ to $[0,1]$ and $\Bbb{R}$ to $[0,1]$, respectively.
Let define $\iota : [0,1]^A\to [0,1]^B$ as follows: for $g\in B$ and $x \in [0,1]^A$, define $$\iota(x)(g) = x(g|_\Bbb{N})$$ (Note that $x$ is a function from $A$ to $[0,1]$, and $\iota(x)$ will be a function from $B$ to $[0,1]$.)
Since each component of $\iota$ is continuous, $\iota$ itself is also continuous. Moreover, since $\Bbb{N}$ is $C^*$-embedded into $\Bbb{R}$, $\iota$ is one-to-one.
For embeddings $h_\Bbb{N}(n) := \langle f(n)\rangle_{f\in A}$ and $h_\Bbb{R}(r) := \langle g(r)\rangle_{g\in B}$ we can show that $$h_\Bbb{R}(n) = \iota(h_\Bbb{N}(n))$$ for all $n\in \Bbb{R}$. Therefore $\iota$ defines a map from $h_\Bbb{N}(\Bbb{N})$ to $h_\Bbb{R}(\Bbb{R})$, and so $\iota$ gives a map from $\beta\Bbb{N}$ to $\beta\Bbb{R}$.