This is a proof in terms of nets. Recall the following facts:

Lemma 1 - A net $(x_{\alpha})_{\alpha \in \mathcal{A}}$ in $\prod_{i \in I}X_i$ converges to $x \in \prod_{i \in I}X_i$ if and only if each coordinate $(x_{\alpha}^i)_{\alpha \in \mathcal{A}}$ converges to $x^i \in X_i$ Lemma 2 - A topological space $X$ is compact if and only if every net in $X$ has a convergent subnet.

Lemma 3 - Every net has a universal subnet.

Lemma 4 - A universal net $(x_{\alpha})_{\alpha \in \mathcal{A}}$ in a compact space $X$ is convergent. (see this entry)

We now prove Tychonoff's theorem.

Proof (Tychonoff's theorem) : Let $(x_{\alpha})_{\alpha \in \mathcal{A}}$ be a net in $\prod_{i \in I}X_i$

Using Lemma 3 we can find a universal subnet $(y_{\beta})_{\beta \in \mathcal{B}}$ of $(x_{\alpha})_{\alpha \in \mathcal{A}}$

It is easily seen that each coordinate net $(y_{\beta}^i)_{\beta \in \mathcal{B}}$ is a universal net in $X_i$

Using Lemma 4 we see that each coordinate net converges, because $X_i$ is compact.

Using Lemma 1 we see that the whole net $(y_{\beta})_{\beta \in \mathcal{B}}$ converges in $\prod_{i \in I}X_i$

We conclude that every net in $\prod_{i \in I}X_i$ has a convergent subnet, so, by Lemma 2, $\prod_{i \in I}X_i$ must be compact. $\square$