Show that $P_n\Rightarrow P$ if and only if $P_n(x_k)\rightarrow P(x_k)$ for every $k\ge$ and $\sum_{k=1}^\infty P(x_k)=1$

Question

Let $C=\{x_1,x_2,...x_k\}$ be a countable set of $\mathbb{R}$. Let $\{P_n\}$ and $P$ be probability measures on $C$. Show that $P_n\Rightarrow P$ if and only if $P_n(x_k)\rightarrow P(x_k)$ for every $k\ge$ and $\sum_{k=1}^\infty P(x_k)=1$.

I have always used this equivalent condition to prove weak convergence assuming its correctness intuitively. For example, it helps when proving Binomial to Poisson convergence etc. I'm not sure how to prove it. Regardless, here is an attempt.

$(\Leftarrow)$ $\int f(x)dP_n(x)=\sum_{k\in\mathbb{N}}f(x_k)P_n(x_k)\overset{n\rightarrow\infty}{\longrightarrow}\underbrace{\sum_{k\in\mathbb{N}}f(x_k)P(x_k)}_{\text{by DCT}}=\int f(x)dP(x)$

$(\Rightarrow)$ I need to find some specific continuous bounded function $f$ for this, but I'm unable to figure it.

Note. I don't want to use other equivalent definitions to prove this as I'm required to prove them via these.