Given a vector valued function $f : \mathbb{R}\to\mathbb{R}^n$, we say that $f$ is continuous at $a$ if $f(a)$ exists and for all $\epsilon > 0$, there exists a $\delta > 0$ such that $d_{\mathbb{R}^n}(f(x), f(a)) < \epsilon$ whenever $d_{\mathbb{R}}(x,a) < \delta$, where $d_{X} : X\times X\to\mathbb{R}$ is the metric you're using on $X$ (here $\mathbb{R}$ and $\mathbb{R}^n$). I'm assuming you're using the standard Euclidean metric on $\mathbb{R}^n$, i.e.
$$
d_{\mathbb{R}^n}(x,y) = \sqrt{\sum_{i = 1}^n(x_i - y_i)^2},
$$
where $x = \begin{pmatrix}x_1\\ \vdots \\ x_n\end{pmatrix}, y = \begin{pmatrix}y_1\\ \vdots \\ y_n\end{pmatrix}$.
With that in mind, suppose each $f_i$ is continuous. Then given $\epsilon > 0$, we can find $\delta_i > 0$ such that $\left|\,f_i(x) - f_i(a)\right| < \epsilon/\sqrt{n}$ whenever $\left|x - a\right| < \delta_i$. Let $\delta = \min_{i = 1,\ldots, n}\delta_i$. Then we have
\begin{align*}
d_{\mathbb{R}^n}(f(x), f(a)) &= \sqrt{\sum_{i = 1}^n\left(\,f_i(x) - f_i(a)\right)^2}\\
&< \sqrt{\sum_{i = 1}^n\left(\epsilon/\sqrt{n}\right)^2}\\
&= \sqrt{\epsilon^2}\\
&= \epsilon,
\end{align*}
whenever $\left|x - a\right| < \delta$, so $f$ is continuous when all of the component functions are continuous. For the reverse direction, you can use a similar idea or you can try to prove the contrapositive: if any $f_i$ is not continuous at $a$, then $f$ is not continuous at $a$ (you'll have to figure out the proper way to negate the statement of continuity, but once you have that, it should be similar to the proof I presented above).
