This proof can be considered a variation of the proof using ultrafilters on $X$. I want mainly to point out that we can avoid transferring the ultrafilters through the projections if we use a slightly more general characterization of compactness using ultrafilters.$\newcommand{\FF}{\mathcal F}\newcommand{\UU}{\mathcal U}$

Definition. Let $X$ be a topological space, $x\in X$, $f\colon M\to X$ be a function and $\FF$ be a filter on $X$. Then we say that $x$ is an $\FF$-limit of $f$ iff for every neighborhood $U\ni x$ we have $$f^{-1}[U]\in\FF.$$

Basically, this definition says that $f^{-1}[U]$ has to be a "big set". (You can compare this with the definition of a limit of a sequence $f\colon\mathbb N\to X$ where $f^{-1}[U]$ has to be a cofinite set, i.e., it belongs to the Fréchet filter.)

Some references concerning this notion can be found in this post: Where has this common generalization of nets and filters been written down?

We can now characterize compactness in the following way

Fact. A topological space is compact if and only if for every function $f\colon M\to X$ and every ultrafilter $\UU$ on $M$ there exists an $\UU$-limit in $X$.

A proof of the "easy" implication can be found, for example, here: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. Proof of the fact that this characterizes compact spaces will probably depend on the facts which were already proven about compact spaces and (ultra)filters at this point.

Proof of Tychonoff theorem. Let $X=\prod\limits_{i\in I} X_i$ be a product of compact spaces. Suppose we have an ultrafilter $\UU$ on $M$ and a function $f\colon M\to X$. Then for each $i\in I$ there exists som $\UU$-limit of $p_i\circ f$ in the compact space $X_i$. Then the point $x$ determined by $p_i(x)=x_i$ is an $\UU$-limit of $f$ in $X$.

(In the proof, we have also used the fact $\FF$-limit in topological product corresponds to pointwise $\FF$-limits for each $i\in I$.)

Proof of Tychonoff's theorem along these lines is given, for example, in Dixmier's _General Topology (zbmath 0545.54001, MR753644) as Theorem 4.3.6.

This proof can be considered a variation of the proof using ultrafilters on $X$. I want mainly to point out that we can avoid transferring the ultrafilters through the projections if we use a slightly more general characterization of compactness using ultrafilters.$\newcommand{\FF}{\mathcal F}\newcommand{\UU}{\mathcal U}$

Definition. Let $X$ be a topological space, $x\in X$, $f\colon M\to X$ be a function and $\FF$ be a filter on $X$. Then we say that $x$ is an $\FF$-limit of $f$ iff for every neighborhood $U\ni x$ we have $$f^{-1}[U]\in\FF.$$

Basically, this definition says that $f^{-1}[U]$ has to be a "big set". (You can compare this with the definition of a limit of a sequence $f\colon\mathbb N\to X$ where $f^{-1}[U]$ has to be a cofinite set, i.e., it belongs to the Fréchet filter.)

Some references concerning this notion can be found in this post: Where has this common generalization of nets and filters been written down?

We can now characterize compactness in the following way

Fact. A topological space is compact if and only if for every function $f\colon M\to X$ and every ultrafilter $\UU$ on $M$ there exists an $\UU$-limit in $X$.

A proof of the "easy" implication can be found, for example, here: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. If presented in some introductory course, the proof of the fact that this characterizes compact spaces will probably depend on the facts which were already proven about compact spaces and (ultra)filters at this point.

Proof of Tychonoff theorem. Let $X=\prod\limits_{i\in I} X_i$ be a product of compact spaces. Suppose we have an ultrafilter $\UU$ on $M$ and a function $f\colon M\to X$. Then for each $i\in I$ there exists som $\UU$-limit of $p_i\circ f$ in the compact space $X_i$. Then the point $x$ determined by $p_i(x)=x_i$ is an $\UU$-limit of $f$ in $X$.

(In the proof, we have also used the fact $\FF$-limit in topological product corresponds to pointwise $\FF$-limits for each $i\in I$.)

Proof of Tychonoff's theorem along these lines is given, for example, in Dixmier's _General Topology (zbmath 0545.54001, MR753644) as Theorem 4.3.6. The whole proof is just a few lines - of course, that is related to the fact that it relies on a lot of things proved before that.

This proof can be considered a variation of the proof using ultrafilters on $X$. I want mainly to point out that we can avoid transferring the ultrafilters through the projections if we use a slightly more general characterization of compactness using ultrafilters.$\newcommand{\FF}{\mathcal F}\newcommand{\UU}{\mathcal U}$

Definition. Let $X$ be a topological space, $x\in X$, $f\colon M\to X$ be a function and $\FF$ be a filter on $X$. Then we say that $x$ is an $\FF$-limit of $f$ iff for every neighborhood $U\ni x$ we have $$f^{-1}[U]\in\FF.$$

Basically, this definition says that $f^{-1}[U]$ has to be a "big set". (You can compare this with the definition of a limit of a sequence $f\colon\mathbb N\to X$ where $f^{-1}[U]$ has to be a cofinite set, i.e., it belongs to the Fréchet filter.)

Some references concerning this notion can be found in this post: Where has this common generalization of nets and filters been written down?

We can now characterize compactness in the following way

Fact. A topological space is compact if and only if for every function $f\colon M\to X$ and every ultrafilter $\UU$ on $M$ there exists an $\UU$-limit in $X$.

A proof of the "easy" implication can be found, for example, here: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. Proof of the fact that this characterizes compact spaces will probably depend on the facts which were already proven about compact spaces and (ultra)filters at this point.

Proof of Tychonoff theorem. Let $X=\prod\limits_{i\in I} X_i$ be a product of compact spaces. Suppose we have an ultrafilter $\UU$ on $M$ and a function $f\colon M\to X$. Then for each $i\in I$ there exists som $\UU$-limit of $p_i\circ f$ in the compact space $X_i$. Then the point $x$ determined by $p_i(x)=x_i$ is an $\UU$-limit of $f$ in $X$.

(In the proof, we have also used the fact $\FF$-limit in topological product corresponds to pointwise $\FF$-limits for each $i\in I$.)

This proof can be considered a variation of the proof using ultrafilters on $X$. I want mainly to point out that we can avoid transferring the ultrafilters through the projections if we use a slightly more general characterization of compactness using ultrafilters.$\newcommand{\FF}{\mathcal F}\newcommand{\UU}{\mathcal U}$

Definition. Let $X$ be a topological space, $x\in X$, $f\colon M\to X$ be a function and $\FF$ be a filter on $X$. Then we say that $x$ is an $\FF$-limit of $f$ iff for every neighborhood $U\ni x$ we have $$f^{-1}[U]\in\FF.$$

Basically, this definition says that $f^{-1}[U]$ has to be a "big set". (You can compare this with the definition of a limit of a sequence $f\colon\mathbb N\to X$ where $f^{-1}[U]$ has to be a cofinite set, i.e., it belongs to the Fréchet filter.)

Some references concerning this notion can be found in this post: Where has this common generalization of nets and filters been written down?

We can now characterize compactness in the following way

Fact. A topological space is compact if and only if for every function $f\colon M\to X$ and every ultrafilter $\UU$ on $M$ there exists an $\UU$-limit in $X$.

A proof of the "easy" implication can be found, for example, here: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. Proof of the fact that this characterizes compact spaces will probably depend on the facts which were already proven about compact spaces and (ultra)filters at this point.

Proof of Tychonoff theorem. Let $X=\prod\limits_{i\in I} X_i$ be a product of compact spaces. Suppose we have an ultrafilter $\UU$ on $M$ and a function $f\colon M\to X$. Then for each $i\in I$ there exists som $\UU$-limit of $p_i\circ f$ in the compact space $X_i$. Then the point $x$ determined by $p_i(x)=x_i$ is an $\UU$-limit of $f$ in $X$.

(In the proof, we have also used the fact $\FF$-limit in topological product corresponds to pointwise $\FF$-limits for each $i\in I$.)

Proof of Tychonoff's theorem along these lines is given, for example, in Dixmier's _General Topology (zbmath 0545.54001, MR753644) as Theorem 4.3.6.