Proof product of components in factors is a component in product topology 
Let $x = (x_1, x_2, .... x_{n})$ be a point in a product space $(Y, \tau_{Y}) = \prod_{i = 1}^{n} (X_{i}, \tau_{i})$. The component $C_{X}(y)$ in a topological space is the union of all connected sets containing $y$.
We need to show:
For a point $x \in Y$(the product topology) $C_{Y}(x) = \prod_{i=1}^{n} C_{X_{i}}(x_{i})$.

I showed the $\supseteq$ inclusion by arguing it is a product of connected factors, so it is connected and it is containing $x = (x_1, x_2, .... x_{n})$, so it is included in the $C_{Y}(x)$ However, I am stuck with the inclusion the other way.
 A: Assume that $\subset$ does not hold, so the connected set $C_Y(x)$ contains a point $z=(z_1,\ldots,z_n)\not\in\prod C_{X_i}(x_i)$. Without loss of generality, $z_1\not\in C_{X_1}(x_1)$. Let $\pi_1$ denote the projection to the first coordinate. By definition of the product topology, $\pi_1$ is continuous and hence carries connected subsets to connected subsets. The set $\pi_1(C_Y(x))$ is thus a connected subset of $X_1$. This is a contradiction, since both $x_1$ and $z_1$ lie in $\pi_1(C_Y(x))$.
A: Here's an attempt at a direct proof:
We want to show $C_{Y}(x) \subset \prod_{i=1}^{n} C_{X_{i}}(x_{i})$
Let $z = (z_1, ..., z_n)\in C_{Y}(x)$
If we can show that $\forall i, z_i\in C_{X_{i}}(x_{i})$, we're done.
Consider the $i$th projection:
$\pi_i(C_{Y}(x)) = \{z_i\in X_i : (z_1, ..., z_n)\in C_{Y}(x)\}\subset X_i$
Note that, $z_i, x_i \in \pi_i(C_{Y}(x))$
Now, since the projection map is continuous, it preserves connectedness, which means that $\pi_i(C_{Y}(x))$ is a connected subset of $X_i$.
But, by definition $C_{X_{i}}(x_{i})$ is the union of all connected subsets of $X_i$ containing $x_i$.  
Therefore, $\pi_i(C_{Y}(x))\subset C_{X_{i}}(x_{i})$
Since this works for all $i$, we have $z = (z_1, ..., z_n)\in \prod_{i=1}^{n} C_{X_{i}}(x_{i})$
