The explicit form of the linear continuous map $\varphi:\mathbb R^I \to \mathbb R$ @Jochen suggested another solution in his comment for this problem. The solution depends on below result.

Let $I$ be a set of indices. We endow $\mathbb R^I$ with the product topology. Let $\varphi:\mathbb R^I \to \mathbb R$ be linear continuous. Then there is a finite subset $I_0$ of $I$ and a collection $(\beta_i)_{i\in I_0}$ of real numbers such that
$$
\varphi((x_i)_{i\in I}) = \sum_{i\in I_0} \beta_i x_i \quad \forall (x_i)_{i\in I} \in \mathbb R^I.
$$


My attempt: For $j\in I$, let $y^j \in \mathbb R^I$ such that $y_i^j = \delta_{ij}$. Let $I_0 := \{j \in I \mid \varphi (y^j) \neq 0\}$.
Assume the contrary that $I_0$ is infinite. WLOG, we assume $I_0 := (j_n)_{n\in \mathbb N}$. Let $(\alpha_n)$ be a sequence in $\mathbb R$. We define a sequence $(z_n)$ by $z_n := \alpha_0 y^{j_0} + \cdots + \alpha_n y^{j_n}$. It's easy to see that $z_n \to z\in \mathbb R^I$ with $z_j = 0$ if $j \notin I_0$ and $z_j = \alpha_n$ if $j=j_n \in I_0$. Because $\varphi$ is linear continuous, $\varphi(z_n) = \alpha_0 \varphi(y^{j_0}) + \cdots + \alpha_n \varphi(y^{j_n}) \to \varphi(z)$. It's possible to construct a sequence $(\alpha_n)$ such that $\varphi(z_n) \to \infty$. This is a contradiction.
Let $x \in \mathbb R^I$ such that $x_i=0$ for all $i \in I_0$. We have
\begin{aligned}
\varphi((x_i)_{i\in I}) &= \varphi \left (\sum_{i\in I_0} x_i y^i + \sum_{i\in I \setminus I_0} x_i y^i  \right) \\
&=  \sum_{i\in I_0} x_i \varphi \left ( y^i \right )  + \varphi \left( \sum_{i\in I \setminus I_0} x_i y^i  \right).
\end{aligned}

Then I'm stuck at proving $\varphi \left( \sum_{i\in I \setminus I_0} x_i y^i  \right) = 0$. Because $I \setminus I_0$ is possibly infinite, I could not use
$$
\varphi \left( \sum_{i\in I \setminus I_0} x_i y^i  \right) =  \sum_{i\in I \setminus I_0} x_i \varphi(y^i)  = \sum_{i\in I \setminus I_0} x_i \cdot 0 = 0.
$$
Could you elaborate on how to get over this difficulty?
 A: Hint for a better proof idea: consider $\phi^{-1}[(-1,1)]$ which is open in $\Bbb R^I$ and contains $0$ (by linearity we know $\phi(0)=0$).
So for some basic subset $\prod_{i \in I} U_i$ we have that all $U_i$ are open neighbourhoods of $0$ in $\Bbb R$, and there is a finite $I_0 \subseteq I$ so that $i \notin I$ implies $U_i=\Bbb R$ and $$\prod_{i \in I} U_i \subseteq \phi^{-1}[(-1,1)]$$
My claim is that this $I_0$ is as required for your result, and points with support outside $I_0$ must map to $0$ under $\phi$.
A: For $i\in I$, let $y^i \in \mathbb R^I$ such that $y^i_j = \delta_{ij}$. Let $U := \varphi^{-1} [(-1, 1)]$. There is a finite subset $I_0 \subseteq I$ and a collection $(U_i)_{i\in I}$ of open subsets of $\mathbb R$ such that $U_i = \mathbb R$ for all $i \in I_0^c := I \setminus I_0$ and that
$$
V := \prod_{i\in I} U_i \subseteq U.
$$
It follows that $n\sum_{i\in I^c_0} x_iy^i \in V$ for all $n \in \mathbb N, x_i\in \mathbb R$. Consequently,
$$
\varphi \left (n\sum_{i\in I^c_0} x_iy^i \right) = n \varphi \left (\sum_{i\in I^c_0} x_iy^i \right) \in (-1, 1) \quad \forall n\in \mathbb N.
$$
So
$$
\varphi \left (\sum_{i\in I^c_0} x_iy^i \right) = 0 \quad \forall x_i \in \mathbb R.
$$
Finally,
$$
\begin{aligned}
\varphi \left (\sum_{i\in I} x_iy^i \right) &= \varphi \left ( \sum_{i\in I_0} x_iy^i + \sum_{i\in I^c_0} x_iy^i \right) \\
&=  \varphi \left ( \sum_{i\in I_0} x_iy^i \right ) + \varphi \left ( \sum_{i\in I^c_0} x_iy^i \right) \\
&=  \varphi \left ( \sum_{i\in I_0} x_iy^i \right )  = \sum_{i\in I_0} x_i\varphi ( y^i) .
\end{aligned} 
$$
This completes the proof.
