For a countable infinite product, each (xi, Ti) is homeomorphic to a subspace of the product "Let($X_i,\tau_i$),i∈N,be a countably infinite family of topological spaces. Prove that each ($X_i,\tau_i$) is homeomorphic to a subspace of $\Pi(X_i,\tau_i)$"
For each j $\ne$ i, fix an element $a_j$ in each j. For each i, define mapping $f_i(x)$: $(X_i,\tau_i$) $\mapsto$ $\Pi(X_i,\tau_i)$ by $f_i(x)$=($a,a_1,a_2$,$a_{i-1}$,x,$a_{i+1}$, $a_{i+2}$,...,$a_n$).
We claim that $f_i(x)$: ($X_i,\tau$’)$\mapsto$($f_i(X_i)$,τ′)is a homeomorphism, where τ′ is the topology subspace topology. Clearly, this mapping is bijective.  Let U∈$\tau_i$ Then$ f_i(U)$={$a_1$}×{$a_2$}×···×{$a_{i-1}$}×U×{$a_{i+1}$}×···×{$a_n$}=($X_1×X_2$×···×$X_{i−1}$×U×$X_{i+1}$×···×$X_n$)∩({$a_1$}×{$a_2$}×···×{$a_{i-1}$}×$X_i$×{$a_{i+1}$×···×{$a_n$})=($X_1×X_2$×···×$X_{i−1}$×
U×$X_{i+1}$×···×$X_n$)$\cap$ $f_i(X_i$))∈τ′ Thus $f_i(U)$ is open and in the subspace topology so we have shown the mapping is open.
Now we just need to show the mapping is continuous. Take any open set in the codomain which is of the form {($U_1×U_2×···×RxRxRXRxR)\cap$ $f_i(X_i)$. Since this is the infinite product topology, all basic open sets must have R for all but finite number of components. Now taking the preimage,$f^{-1}_i[(U_1×U_2$×...x X x X x X x X)∩$f_i(X_i)]$=$f^{-1}_i$($U_1×U_2$×...x X x X)$\cap$ $f^{-1}_i$=$f^{−1}_i$($U_1×U_2×...x X x X xX$)$\cap$($X_i$).
Now, if $a_j \notin$ $U_i$ then $U_i$ interection with $X_i$ is empty. Otherwise, the intersection is $U_i$ which is open in $T_i$ Either way, we get an open set, so the mapping is continuous.
One area of concern is whether it is possible to fix an element from each $X_i$ when we have an infinite set. I know it is possible for finite number of sets.
 A: You'll have to assume $X:=\prod_{i \in I} X_i$ is non-empty, or equivalently that all $X_i$ are non-empty. The Axiom of Choice then allows us me to pick a point $a \in \prod_{i \in I} X_i$ and I'll fix that point from now on.
Then for any $j \in I$ we define $e_j: X_j \to X$ by
$$\pi_i(e_j(x))= \begin{cases} x & i=j\\
                               a_i & i \neq j\\
\end{cases}$$
and note that $e_j$ is continuous as all $\pi_i \circ e_j$ are either a constant map (so continuous) or the identity on $X_j$ (also continuous). The universal property of products strikes again. No separate proof needed.
$e_j$ is 1-1: $x \neq x'$ implies $e_j(x)_i \neq e_j(x')_i$ so $e_j(x) \neq e_j(x')$.
To show its en embedding it is enough to it has a continuous inverse $e_j[X_j] \to X_j$ and this inverse is $\pi_j\restriction_{e_j[X_j]}$, which is clearly continuous and the required inverse. Don't overcomplicate things.
The idea of the map was OK though. There is no doubt in my mind that we can fix such a point. I believe in AC, as do most topologists. Not going to give up Tychonoff's theorem after all...
