I suppose the problem is with a lower semicontinuous function that is not bounded below. If the function is bounded below (without loss of generality $f \geqslant 0$) but not above, one can approximate the bounded function $\tilde{f}(x) = \dfrac{f(x)}{1+f(x)}$ by an increasing sequence $g_n$, cap each $g_n$ at $1-\frac1n$, $h_n(x) = \min \left\{ g_n(x),\, \frac{n-1}{n}\right\}$, and use
$$k_n(x) = \frac{h_n(x)}{1-h_n(x)}$$
for the approximating monotonic sequence of continuous functions.
For $f$ not bounded below, if we find a continuous $g \leqslant f$, we can reduce the approximation to the above, $h = f-g \geqslant 0$ is lower semicontinuous, and if $k_n \uparrow f-g$, then $k_n+g \uparrow f$.
So it remains to find a (finitely valued) continuous $g \leqslant f$. Since a lower semicontinuous function attains its minimum on any compact subset of $\mathbb{R}$, we have no problem finding a continuous function $g_{a,b}$ on $[a,b]$ that is a lower bound of $f$ there (for example a constant function). Then, using a partition of unity, we can glue those lower bounds together to obtain a global continuous $g\leqslant f$.
For example, let
$$\psi(x) = \begin{cases}\qquad 0 &, \lvert x\rvert \geqslant \frac34\\
2\left(x+\frac34\right) &, -\frac34 \leqslant x \leqslant -\frac14\\
\qquad 1 &, \lvert x\rvert \leqslant \frac14\\
2\left(\frac34-x\right) &, \frac14 \leqslant x \leqslant \frac34 \end{cases}$$
Then the translates of $\psi$ by integers form a continuous partition of unity,
$$\sum_{k\in\mathbb{Z}} \psi(x-k) \equiv 1,$$
and letting
$$c_k := \min \{ f(x) : x \in [k-1,\,k+1]\}$$
we obtain a continuous function $g \leqslant f$ by setting
$$g(x) = \sum_{k\in\mathbb{Z}} c_k\cdot \psi(x-k).$$
Recall that a function $f$ is upper-semicontinuous at $x_0 \in \mathbf R^N$ iff
$$ \limsup_{x\to x_0} f(x) \le f(x_0) $$
Now let $K \subseteq \mathbf R^N$ be a compact subset, $m := \sup_{x \in K} f(x)$. For every $n \in \mathbf N$, choose $x_n \in K$ such that $f(x_n) \ge m - \frac 1n$. As $K$ is compact, some subsequence $(x_{n_k})$ converges, say $x_{n_k} \to x_0$. Then, by semi-continuity,
$$ m \ge f(x_0) \ge \limsup_{x \to x_0} f(x) \ge \limsup_{k \to \infty} f(x_{n_k}) \ge \lim_k m - \frac 1{n_k} = m $$
Hence $m = f(x_0)$ and $f$ attains its maximum on $K$. Now use the same argument as for continuous functions.
Best Answer
I came across this looking for a wrong theorem. Sorry if it is too late. I would avoid taking sups because they may not preserve smoothness. But if you know an increasing sequence of continuous functions that converge to your lsc function, you may obtain smooth ones by removing $2^{-n}$ to the current function, approximate it within $2^{-n-2}$ by a smooth function (but in the whole space $R^d$ a brutal convolution will not work, I don't know a better way than using partitions of unity before you convolve). Anyway, your new sequence is smooth and still increasing, and converges to the same limit. Agreed, this will not be nonnegative if your initial function $f$ was zero somewhere. For that case I am afraid I see no way to avoid doing this by hand, working on the open set where $f > 2^{-n}$, doing the same sort of thing as above there, and gluing by hand in the remaining region. (Sorry, did not spend too much time).