Continuity in $\mathbb R$ results in continuity in $\mathbb R^2$; Proof? During studying of proof of some other theorem, I faced with the claim (without proof): 

since $f(x,t)$ and $g(x,t)$ are continuous functions [$f,g:\mathbb R^2 \rightarrow \mathbb R$] thus the function $F:\mathbb R^2 \rightarrow \mathbb R^2$, $F(x,t)=(f(x,t),g(x,t))$ is continuous. 

It seems obvious but having rigor-obsession, would someone please guide me how to prove it rigorously? I don't know how to start because I know one-dimensional and one parameter $\epsilon-\delta$ proof of continuity.   
Thanks a lot. 
EDIT - especially, What does it mean a two-parameter function to be continuous?
 A: (I'm probably going into topics you haven't covered, so bear with me.)
Have you heard of a "topology"? A topology is probably the most general "useful" object for doing analysis. Basically, a topology on a set $S$ (it can be any general set) is a set of subsets of $S$ satisfying three properties, which you can research yourself if you're interested. These sets are the analogue of open subsets of $S$. That is, when you define a topology, you manually define which sets are open, and their complements are considered the closed sets.
There's no concept of distance defined here, but remarkably, it still allows us to talk about limits, continuity, compactness, closures, interiors, and a host of other analytic concepts. The real line $\mathbb{R}$ has its own topology, which is just the set of open sets (i.e. sets which have the property that every point within them have an open interval around the point that are also contained in the set).
More generally, there's another concept of a metric space ($\mathbb{R}$ is also a metric space). Basically, it's a set with a function that measures some kind of "distance" between points, which also satisfies various axioms. Every metric space has a topology of open sets as well.
Anyway, as it turns out, if you have two topological spaces $X$ and $Y$, then there's a natural topology on the cartesian product $X \times Y$ of the sets. This topology is called the "product topology". So, it allows us to talk about things like continuity on $X \times Y$, and we do so in terms of continuity on $X$ and $Y$. If $X$ and $Y$ have a distance function (a.k.a. a "metric") defined on them, then there is at least one metric defined on $X \times Y$ that generates the product topology. Basically, we can define a distance on $X \times Y$ so that all the limits, continuity, etc that we get from the metric agrees with the corresponding limits, continuity, etc from the product topology. However, there will always be infinitely many such metrics on $X \times Y$.
That's what's happening here: we're examining the continuity of a function on $\mathbb{R} \times \mathbb{R}$, and we need to do so in terms of continuity on $\mathbb{R}$. The standard metric on $R$ is given by $(x, y) \mapsto |x - y|$. From this, we can induce several metrics on $\mathbb{R}^2$. For example,
\begin{align*}
((x_1, y_1), (x_2, y_2)) &\mapsto \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\
((x_1, y_1), (x_2, y_2)) &\mapsto |x_1 - x_2| + |y_1 - y_2| \\
((x_1, y_1), (x_2, y_2)) &\mapsto \max \lbrace |x_1 - x_2|, |y_1 - y_2| \rbrace
\end{align*}
are all metrics that all generate the product topology. You can use any one of them to talk about continuity. So, just from these metrics alone, we have three different, but all equivalent, notions of continuity of a function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ at a point $(a, b)$:
\begin{align*}
\forall \varepsilon > 0, \exists \delta > 0 &: 0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta \implies |f(x, y) - f(a, b)| < \varepsilon \\
\forall \varepsilon > 0, \exists \delta > 0 &: 0 < |x - a| + |y - b| < \delta \implies |f(x, y) - f(a, b)| < \varepsilon \\
\forall \varepsilon > 0, \exists \delta > 0 &: 0 < \max \lbrace |x - a|, |y - b| \rbrace < \delta \implies |f(x, y) - f(a, b)| < \varepsilon.
\end{align*}
We also get three separate notions of continuity of a function $h : \mathbb{R}^2 \rightarrow \mathbb{R}^2$, which will essentially be the same as the above definitions, but with $|f(x, y) - f(a, b)|$ replaced by an equivalent expression involving one of the three metrics. For example, if we denote $h(x, y) = (p(x, y), q(x, y))$, then one such definition of continuity could be
$$\forall \varepsilon > 0, \exists \delta > 0 : 0 < |x - a| + |y - b| < \delta \implies |p(x, y) - p(a, b)| + |q(x, y) - q(a, b)| < \varepsilon.$$
So, with that in mind, let's actually answer the question! Suppose $f, g : \mathbb{R}^2 \rightarrow \mathbb{R}$ are continuous, and let $h(x, y) = (f(x, y), g(x, y))$. We wish to prove $h$ is continuous at any point $(a, b)$.
Fix $\varepsilon > 0$. By continuity of $f$ and $g$ respectively, there exist some $\delta_f, \delta_g$ such that
\begin{align*}
0 < |x - a| + |y - b| < \delta_f &\implies |f(x, y) - f(a, b)| < \frac{\varepsilon}{2} \\
0 < |x - a| + |y - b| < \delta_g &\implies |g(x, y) - g(a, b)| < \frac{\varepsilon}{2}.
\end{align*}
Let $\delta = \min \lbrace \delta_f, \delta_g \rbrace$. Then,
$$0 < |x - a| + |y - b| < \delta \implies |f(x, y) - f(a, b)| + |g(x, y) - g(a, b)| < \varepsilon,$$
which proves $h$ is continuous, as required.
I hope that helps. :-/
A: Try to use a good norm, for instance$||F(x,t)-F(x_0,t_0)||=\max\{||f(x,t)-f(x_0,t_0)||,||g(x,t)-g(x_0,t_0)||\},$
and then use the continuity of the functions $f, g$. And remember that all norms in finite dimensional vectors spaces are equivalents. Then continuity in one is continuity in another.
