A bounded function in $\Bbb R$ with closed graph is continuous.

Another way to think about this (not following the OP's approach, but using compactness in a slightly different way):

To show that $f$ is continuous, it's enough to show that the restriction $f_{|I}: I \to \mathbb R$ is continuous for each closed interval $I \subset \mathbb R$; this is what we'll do.

Let $\Gamma_f \subset \mathbb R^2$ denote the graph of $f$, which is closed by assumption. Then $\Gamma_{f_{|I}} := \Gamma_f \bigcap (I \times \mathbb R)$, which is the graph of $f_{|I}$, is closed in $I\times \mathbb R$. The assumption that $f$ is bounded shows that it is also bounded, and so it is a compact subset of $I\times \mathbb R$.

Now it is a general property of graphs that the projection $\Gamma_{f_{|I}} \to I$ given by $(x,y) \mapsto x$ is continuous and bijective. A continuous bijection between compact topological spaces is necessarily a homeomorphism. Taking the inverse homeomorphism we obtain a homeomorphism $I \to \Gamma_{f_{|I}}$, which when composed with the (continuous) projection $\Gamma_{f_{|I}} \to \mathbb R$ defined by $(x,y) \mapsto y$, gives the function $f_{|I}: I \to \mathbb R$, but now realized as a composite of continuous functions. This shows that $f$ is continuous.

(We also see that we don't need to assume that $f$ is bounded, but just that $f$ is bounded when restricted to bounded subsets of $\mathbb R$ --- a property that every continuous function on $\mathbb R$ satisfies; of course, this is also clear with the OP's proof strategy (as completed in the other answers). )

Very good so far. To finish it, assume that $\bigl(f(x_n)\bigr)$ didn't converge to $f(x)$. Then there would be an $\varepsilon > 0$ and a subsequence $(x_{n_k})$ of $(x_n)$ with $\lvert f(x_{n_k}) - f(x)\rvert \geqslant \varepsilon$ for all $k$. Apply your argument to the sequence $(y_k)$ given by $y_k = x_{n_k}$ to obtain the contradiction that $f(y_{k_m}) \to f(x)$ for some subsequence of $(y_k)$, although by construction $\lvert f(y_k) - f(x)\rvert \geqslant \varepsilon$ for all $k$. So the initial assumption is untenable, and we conclude that $f(x_n) \to f(x)$. Since the sequence $(x_n)$ converging to $x$ was arbitrary, $f$ is continuous at $x$. Since $x$ was arbitrary, $f$ is globally continuous.

Here's a summary of the ideas presented above.

If $\lim_{n\to\infty} x_n =x$, yet $f(x_n)$ does not converge to $f(x)$, WLOG, $\limsup_{n\to\infty} f(x_n) > f(x)$ (otherwise replace by $-f$), then there exists a subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} f(x_{n_k}) = \limsup_{n\to\infty} f(x_n)> f(x)$. Since $G(f)$ is closed, $(x,\limsup_{n\to\infty} f(x_n))\in G(f)$, in contradiction to $G$ being the graph of a function.