The question was to prove:
$P_k:(x_1, x_2, \ldots , x_n) \to x_k,\ k \in\{ 1, 2,\cdots n\}$ are continuous functions.
The professor proved this by $$d (x, y) = \sqrt{\sum\limits_{i=1}^n(x_i-y_i)^2} \to 0\implies |x_k - y_k| \le d (x, y) \to 0$$ where $d (x, y)$ is the function for finding the distance between $x$ and $y$. I don't really get the logic of the professor.
Can anyone explain what the professor is saying or show a different, easier proof?
$d(x,y)=\sqrt{(x_1-y_1)^2+\cdots +(x_n-y_n)^2}\ge\sqrt{(x_k-y_k)^2}=|x_k-y_k|$ So given an $\epsilon$ you can take $\delta=\epsilon$, since if you have $d(x,y)< \delta$ then $|x_k-y_k|< \delta=\epsilon$. So the projection function is continuous.
The previous answer was using the definition of continuity in real analysis, assuming that you probably have the topology of $\mathbb{R^n}$ defined as the topology generated by open balls (since open balls are defined with the distance function). This definition of continuity is equivalent to the topological one for spaces like $\mathbb{R^n}$.
Maybe you want to use the more general definition of continuity given for topological spaces: a function $f:X\to Y$ between two topological spaces is continuous if given any open set $V\subseteq Y$, its inverse image $f^{-1}(V)\subseteq X$ is an open set (in the topology of $X$). If that's the case then you probably have the topology of $\mathbb{R^n}$ defined as the product topology of $\mathbb{R}$ with itself $n$ times (the definition saying that a set of $\mathbb{R^n}$ is open if it's the cartesian product of open sets in $\mathbb{R}$). Let's see then if we have continuity with this definition. Take any open set $V\subseteq \mathbb{R}$. You can check that its inverse image by the projection function $P_k$ is $P_k^{-1}(V)=\mathbb{R}\times\cdots\times V\times\cdots\mathbb{R}$, where $V$ is in the $\textit{k-th}$ position ($P_k$ is projecting the $\textit{k}$ coordinate of that product). This is an open set by definition, since is the cartesian product of open sets, so the function is continuous.