Does this infimum tend to infinity? Let $f:\mathbb{R}^2\to\mathbb{R}$ be a continuous function satisfying $$\lim _{x\to +\infty}f(x,y)=+\infty\quad\text{for each fixed }y\in\mathbb{R}.$$ Further, let $\mathcal{I}\subset\mathbb{R}$ be a compact set and define $g:\mathbb{R}\to\mathbb{R}$, $g(x)=\inf_{y\in\mathcal{I}}f(x,y)$
Is it true that $$\lim _{x\to +\infty}g(x)=+\infty$$
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 A: My sincere apologies for my earlier mistake; the answer turns out to be NO after all. A counterexample is given by $f(x,y) = |y-e^{-|x|}|e^{x^2}$ and $I = [0,1]$ in this answer.

We can read below that I asserted without proof that the set:
$$I_M = \{y \in I \mid \forall x > M: f(x,y) > N \}$$
is open in $I$. After close scrutiny, it became clear that while it may be that $0 \in I_M$ for an appropriate $M$, there would always be $y$ close to $0$ that aren't — e.g. $y = e^{-2M}$ has $f(2M, y) = 0$.
The governing principle here is that we are looking at a set of the form:
$$S(U,V) = \{y: y \times U \subseteq V\}$$
with $U, V$ open -- namely, $U = (M, \infty)$ and $V = f^{-1}[(N,\infty)]$, the preimage of $(N, \infty)$ under $f$ (open by continuity of $f$). In general, these sets are not open. For, let $U = \Bbb R$ and $V$ given by:
$$V := \{(x,y) \mid x \le 0 \text{ or } y < \frac 1x \}$$
As you can see, every point of the form $(0,y)$ has $y < \dfrac 1x$ for $0 < x < \dfrac1y$, from which we can see that $V$ is open. However, it is clear that $S(U,V) = \Bbb R_{\le 0}$ is not open.

The argument below is fallacious, as exemplified by $f(x,y) = |y-e^{-|x|}|e^{x^2}$. $I_M$ is not necessarily open.

Yes. 
The assumption on $f$ translates to the following:
$$\forall y \in I: \forall N > 0: \exists M > 0: \forall x > M: f(x,y) > N$$
On the other hand, the desired conclusion is:
$$\forall N > 0: \exists M > 0: \forall x > M: g(x) > N$$
which is to say (using that $I$ is compact to know that $\inf\limits_{y \in I} f(x,y)$ is attained):
$$\forall N > 0: \exists M > 0: \forall x > M: \forall y \in I: f(x,y) > N$$

With our goal clear, let us fix $N > 0$. For $M \in \Bbb N$, define $I_M$ by:
$$I_M = \{y \in I \mid \forall x > M :f(x,y) > N\}$$
Then since $f$ is continuous, $I_M$ is an open set (respective to $I$). Moreover, the assumption on $f$ ensures that for every $y \in I$, there is some $M \in \Bbb N$ such that $y \in I_M$.
Therefore, $I = \bigcup\limits_{M = 1}^\infty I_M$. Moreover, $I_M \subseteq I_{M+1}$. By compactness† of $I$, finitely many $I_{M_i}$ suffice; by the containment relation between them, it follows that $I = I_M$ with $M = \max_i M_i$.
Recalling the definition of $I_M$, we have now established that:
$$\forall x >M: \forall y \in I: f(x,y)> N$$
meaning we've found the $M$ we were looking for. 
Hence we obtain our desired conclusion: $$\lim_{x\to\infty} g(x) = +\infty$$

†: Compactness is read here as "from every cover by open sets, a finite subset of the open sets already covers the space". That this is equivalent to the definition "closed and bounded" for $\Bbb R$ is the Heine-Borel Theorem.
A: The short answer is yes, because the limit on $f(x,y)$ is insensitive to the value of $y$, thus constraining $y$ to some compact subset doesn't affect the limit.
More rigorously:
$$\lim _{x\to +\infty}f(x,y)=+\infty\quad y\in\mathbb{R} \text{ and } \mathcal{I}\subset \mathbb{R}\implies \lim _{x\to +\infty}f(x,y)=+\infty\quad y\in \mathcal{I}\subset\mathbb{R}$$
But, $\forall x.g(x)\in\{f(x,y):y\in \mathcal{I}\}\implies \forall x.\exists y\in \mathcal{I}: g(x)=f(x,y) \implies \{\lim \limits_{x\to +\infty}g(x):x\in \mathbb{R}\}\subset\{\lim \limits_{x\to +\infty}f(x,y):y\in \mathcal{I}\}$
But, $\{\lim \limits_{x\to +\infty}f(x,y):y\in \mathcal{I}\} = \{+\infty\} \implies \{\lim \limits_{x\to +\infty}g(x):x\in \mathbb{R}\}=\{+\infty\} \quad \square$
The key here is that $f(x,y)$ is continuous and its pointwise limits are all $+\infty$. If we allowed this function to be discontinuous, then we could define an infinite subset of $y\in\mathcal{I}$that, while individually tending to $+\infty$ do so at progressively higher values of $x$ so that $g(x) = 0 \quad \forall x\in \mathbb{R}$. In other words, for every $x$, we can find a value of $y$ that makes the function $f(x,y)=0$, so the limit will be $0$ and not $+\infty$.
Clarification
In the above, it was not assumed that $y$ is fixed. I imagined a set of "strips" in the X-Y plane (unbounded in X). In order for your condition to not hold, each line along the x-axis in such a strip must diverge. This eliminates cases where all values of the function for a particular $y$ and all x's converge to some finite number, then the function"pops" up to infinity at $x=\infty$. Therefore, $g(x)$ is a monotonically increasing function with no upper bound.
