Let $X,Y,Z$ be topological spaces, and $f: X \times Y \to Z$ is continuous. Fix $x \in X$ and define $f_x: Y \to Z$ as $f_x(y) = f(x,y)$. Prove that $f_x$ is continuous.
Fix an $y \in Y$ and consider and open set $O \subset Z$ with $f_x(y) = f(x,y) \in O$. We want to show that there is some open set $P \in Y$ with $y \in P$ and $f_x(P) \subset O$. By the continuity of $f$, we know that $P' = f^{-1}(O)$ is open in $X \times Y$, and contains the point $(x,y)$. But I am not sure if this is useful.
Recall that the product topology is generated by the basis $$\{U \times V \ | \ U \subseteq X, V \subseteq Y \text{ are open}\}.$$ Define $g : Y \rightarrow X \times Y$ by $g(y) := (x,y)$. We note that $f_x := f \circ g$. If we can show that $g$ is continuous, we are done (composition of continuous functions is continuous). To show that $g$ is continuous, take an element in the basis for the topology, i.e. $U \times V$ with $U \subseteq X$ and $V \subseteq Y$ open. Then $g^{-1}(U \times V) = \begin{cases} V \text{ if } x \in U \\ \varnothing \text { if } x \notin U\end{cases}$ is open, so the map is continuous.
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The map $i_x: Y \to Y \times Z$ given by $i_x(y)=(x,y)$ is continuous, as $\pi_X \circ i_x \equiv x$ (constant so continuous) and $\pi_Y \circ i_x=1_Y$ (identity so continuous), and the mapping criterion for maps into the product applies.
Finally $f_x= f \circ i_x$ is a composition of continuous maps, hence continuous.