On Wikipedia it states that for any Tychonoff space $X$ there is a natural embedding into $[0,1]^{C(X,[0,1])}$. I assume this embedding is $\iota(x)(f)=f(x)$. I am able to prove that $\iota$ is continuous and injective. However, I am not able to see why $\iota$ is open onto its image.
I have one more question about this embedding, since Wikipedia also states this is the Stone Čech compactification. However I do not see why $\iota(X)$ is dense. Wouldn't the Stone Čech compactification be $\beta X=\overline{\iota(X)}$?
Showing $l^{-1}$ is continuous is equivalent to showing $l(x_\alpha) \to l(x)$ implies $x_\alpha \to x$ for any net $(x_\alpha)$. Suppose $x_\alpha \not \to x$. Then there is some neighborhood, $U$, of $x$ such that $x_\alpha$ frequently lies outside $U$. By definition of a Tychonoff space, there is some continuous function $f : X \to [0,1]$ with $f(x) = 1$ and $f|_{U^c} \equiv 0$. For this $f$, it holds that $f(x_\alpha) \not \to f(x)$. In other words, $(l(x_\alpha))(f) \not \to (l(x))(f)$, which in particular means $l(x_\alpha) \not \to l(x)$ (since $[0,1]^{C(X,[0,1])}$ is endowed with the topology of pointwise convergence).
As for the second part of your question, you are correct: $\beta X = \overline{l(X)}$.